Topic 1 Quantitative Chemistry

Topic 1: Quantitative chemistry

Stoichiometry is the study of the relationship or ratio’s between the amounts of reactants and products in a chemical reaction.

Stoichiometric amounts or ratio’s are the amounts of reactant or product as given by the coefficients in the balanced equation.

The following laws and ideas underpin our knowledge about the relationships between chemicals involved in reactions (see PowerPoint):

1. Law of Conservation of Mass: Lavoisier (18th Century) concluded, from studying reactions involving oxygen, that during any chemical reaction no atoms were destroyed.

2. Law of Definite Proportions: analytical tests showed that any compound consists of the same elements in definite proportions e.g. in every water molecule 88.88% of its mass is oxygen.

3. Law of Combining Volumes of Gases: Gay-Lussac (1803) concluded that when gases react they do so in whole number ratios in terms of volume e.g. 3 volumes of hydrogen react with 1 volume of nitrogen to form 2 volumes of ammonia.

4. Avogadro’s Theory (1811): Avogadro stated that equal volumes of gases must have equal number of particles (provided temperature and pressure are the same)

5. Law of Multiple Proportions: in 2 different compounds of the same elements (e.g. CO and CO2) if the mass of 1 element is the same (e.g. carbon in CO and CO2) than the ratio of the mass of the other element (in our example oxygen) in each compound must be a whole number.

1. 1 The mole concept and Avogadro’s constant

As particles are extremely small chemists measure amounts of substance using a quantity called the MOLE which stands for 6.02 x 1023 a number which is know as Avogadro’s constant.

So 1 mole of a substance = 6.02 x 1023 units of that substance; these units can be atoms, molecules, ions, electrons or formula units.

1 mole = 6.02 x 1023 atoms, ions, molecules, electrons, …Determining number of moles and the number of particles

Use the expression: number of particles number of moles = 6.02 x 1023

Topic1notes 12.5 hours Page 1 of 31

1.1.1 Apply the mole concept to substances.1.1.2 Determine the number of particles and the amount of substance (in moles).

Molar ratios in compounds or formula units:

Examples:

in 1 mole of Fe2 O3 there are 2 moles of iron atoms and 3 moles of oxygen;

in 3 moles of Na2CO3 there are 3 moles of CO32- and 6 moles of Na+ ;

in 2 moles of (NH4)3 PO4 there are 6 moles of NH4+ and 12 moles of hydrogen atoms and so on…

if there are 2 x 1022 C-atoms in a sample of ethane how many hydrogen atoms are there in that same sample?

if you have 6 x 1022 atoms of hydrogen, how many molecules of ammonia do you have?

Exercises

1. How many moles of oxygen are there in 2 moles of Al2O3?

2. How many moles of NO3- are there in 1 mole of Ga(NO3)3?

3. What is the total number of atoms of 0.05 moles of NH4NO3?

1.2. Formulas

Relative isotopic mass

The mass of a specific isotope of an element compared with 1/12 th of the mass of the most common carbon isotope which itself has been given a relative mass of 12. As this is a relative quantity no unit is needed.

actual mass of isotope relative isotopic mass = 1/12th of the mass of a C-12 atom

Relative atomic mass ( Ar)

The relative atomic mass is the average mass of all isotopes of an element again compared to the same standard which is 1/12 th of the mass of one 12C atom. It is calculated as shown below:


1.2.1 Define the terms relative atomic mass (Ar) and relative molecular mass (Mr).1.2.2 Calculate the mass of one mole of a species from its formula.1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass and molar mass.1.2.4. Distinguish between the terms empirical formula and molecular formula.1.2.5 Determine the empirical formula from the percentage composition or from other experimental data.1.2.6 Determine the molecular formula when given both the empirical formula and experimental data.

(isotopic mass 1 x %) + (isotopic mass 2 x %) + (isotopic mass 3 x %) Ar = 100

It is the relative atomic mass which is found in the Periodic Table or data booklet.

Example:

The average magnesium atom has a relative atomic mass of 24 which means that a magnesium atom is 2 times heavier than a carbon atom which has a mass of 12 and 24 times heavier than a hydrogen atom which has a mass of 1.Complete the following table:

element isotopes abundancerelative

atomic mass

CuCu-63 (relative isotopic mass = 63)Cu-65 (relative isotopic mass = 65)

69.09%30.91%

EuEu-151Eu-153

151.9

AgAg-106.9041Ag-108.9047

48.17%

Relative molecular mass ( Mr )

The average mass of a molecule compared to 1/12 the of the mass of one 12C atom.

To be able to calculate the Mr correctly, the formula of the compound must be correct and the relative atomic masses of the elements in the compound need to be used.

Examples:

relative molecular mass of H2O is 18 because (2 x 1.01) + (1 x 16.00) = 18.02

relative molecular mass of H2SO4 is 98 because (2 x 1.01) + 32.06 + (4 x 16.00) = 98.08

Relative formula mass

The Mr only really applies to molecules (a group of covalently bonded atoms); the concept of formula mass is used for giant structures like ionic substances (eg sodium chloride, copper sulphate) and macromolecules (eg silicon dioxide)

Examples:

relative formula mass of NaOH is 40 because 22.99 + 16.00 + 1.01 = 40.00

relative molecular mass of Al2(SO4)3 is because (2 x 26.92 ) + (3 x 32.06) + (12 x 16.00) = 342.02

Molar mass

How do we measure 1 mole of a substance?


The mass of 1 mole of a substance = molar mass which is measured in g mol-1.

The molar mass is the mass of 6.02 x 1023 units of a substance; it is the relative atomic, molecular or formula mass but expressed in grams as the molar mass is an absolute mass.

Problems involving moles, mass and molar mass

Below are examples of the type of calculations you should be able to solve using the mole concept

Calculate the number of moles (using molar mass) when you are given the mass of the substance mass of substance number of moles = molar mass

Calculate actual masses of atoms or molecules

Find the mass of 1 molecule of ethanol, C2H5OH

The only actual mass you have is the molar mass of ethanol which can be calculated using a periodic table

divide by 6.02 x 1023

Calculate number of atoms in a sample

Find the number of carbon atoms in 0.1 mole of C2H5OH.

multiply multiply by 6.02 x 1023 by subscript of atom in formula

Calculate the number of particles from a mass

Which sample contains the greatest number of particles? 5g of NH3 or 5 g of H2O or 5 g of CaO?


molar mass mass of 1 atom/molecule

number of atomsnumber of moleculesnumber of moles

number of particlesnumber of molesgiven mass

divide by molar mass multiply by 6.02 x 1023

Exercises

1. Find the mass of 1 molecule of CH3COOH?

2. Find the mass of 1 atom of Br?

3. Find the mass of 0.05 moles of NH4NO3.

4. Which sample contains the greatest number of molecules?

10g of C2H4 10g of C2H6 10g of C6H6

5. Which sample contains the lowest number of molecules?

2g of CH4 2g of NH3 2g of CO2

6. How many molecules are there in 0.23 g of ethanol?

7. Which one of the following has the greatest mass?

1.0 x 10-3 moles of CH3COOH 1.2 x 1021 atoms of argon 2.4 x 1020 molecules of C8H18

8. How many glucose molecules are there in 5.23 g of C6H12O6?

Using the mole concept and molar ratios for analytical purposes: empirical and molecular formulae.

Empirical formula

To determine the empirical formula (=the most simple molar ratio) of a compound, a chemist needs to find out how much (a mass or a percentage) of each element there is in a certain mass of the compound. This raw data (how much there is of an element in a compound) can be obtained in various ways:

Combustion analysis (usually used with organic compounds)

This involves combusting a known amount of the organic compound and finding the masses of carbon dioxide and water (=raw data). Using the percentage composition of hydrogen and carbon in both compounds allows the chemist to calculate the starting amount of hydrogen and carbon in the compound.

Direct determination (used for binary compounds)

Involves reacting the known mass of one element with a second element, measuring the mass of the compound and then finding the mass of the second element.

Decomposition (usually used to find the formula of a hydrated salt )

Decompose (usually using heat) a sample with a known mass, measure the mass of the final product and use this data to


From this mass analysis of a compound the amount in mass of each element or the percentage composition of the compound can be determined; from this information the most simple whole number molar ratio can be found and this then gives us the empirical formula of the compound.

Percentage composition of a compound

Experimental analysis usually results in the amount of an element in a compound being expressed as a percentage composition

Calculate

1. Percentage by mass of C in CH4, C2H6 and C2H4

2. Percentage by mass of Cu in CuSO4, CuO and Cu2O

Example calculation of empirical formula: analysis of methane:

elements in compound carbon hydrogen

mass in grams (or %) 47.98 (74.9%) 16.02 (25.1%)

number of moles 47.98/12 = 3.99 mole 16.11/1 = 16.02 moles (74.9/12 = 6.24 moles) (25.1/1 = 25.1 moles) ratio of moles 4 (6.25) 16 (25)

most simple ratio of moles 1 4

empirical formula: CH4

Molecular formula

The actual formula of the compound i.e. the molecular formula - which shows the actual number of atoms and ions present in a compound - is always a multiple of the empirical formula both in terms of mass and particles; it is up to the researcher to determine how many times heavier than the empirical formula the actual molecule is.

Strictly speaking we can not use the term molecular formula when dealing with giant structures as they do not consist of separate molecules; in that case any formula is always an empirical formula indicating the most simple ratio of ions or atoms e.g. in the case of silicon dioxide, SiO2 , this formula tells us that there are twice as many oxygen atoms within the lattice as silicon atoms. The formula CuCl2, tells that there are twice as many Cl ions in the giant ionic lattice as copper ions.


molar mass of element x coefficient in compound% composition = x 100 molar mass of compound

There are many possible molecular formulae that can be derived from one empirical formula. However, if the molecular formula is known, the molecular formula can be obtained.

Example: Calculate the molecular formula of a compound with a molecular mass of 84g and an empirical formula of CH2.

Answer: mass empirical formula: 14g molar mass of formula: 84g ratio of molar mass/empirical mass = 84/14 = 6 molecular formula = (CH2 ) x 6 = C6H12

Exercise: Naphthalene, best known as ‘mothballs’, is composed of carbon (93.71%) and hydrogen (6.29%). If the molar mass of the compound is 128g, what is the molecular formula of naphthalene?

Answer: carbon hydrogen

mass of element 93.71 6.29

number of moles 93.71/12 = 7.8 6.29/ 1 = 6.29

DO NOT ROUND UP OR DOWN AT THIS TIME !!!!!!!!!

most simple ratio 7.8/ 6.29 = 1.25 6.29/6.29 = 1

AGAIN DO NOT ROUND UP OR DOWN !!!!!!

lowest whole number ratio 5 4

empirical formula: C5H4

ratio molecular formula /empirical formula: 128g/ 64g = 2

molecular formula = 2 x C5H4 = C10H8

Exercises:

Calculate

i. The empirical formula

ii. The molecular formula

For each of the following substances:

1. Water (one mole = 18g) that contains 11.1% of hydrogen and 88.9% of oxygen.

2. Ammonia (one mole = 17g) that contains 82.4% of nitrogen and 17.6% of hydrogen.

3. Potassium carbonate (one mole = 138g) that contains 56.5% of potassium, 8.7% of carbon and 34.8% of oxygen.


4. Hydrazine (one mole = 32g) that contains 87.5% of nitrogen and 12.5% of hydrogen.

5. Glucose (one mole = 180g) that contains 40.0% of carbon, 6.67% of hydrogen and 53.3% of oxygen.

6. Ethane (one mole = 30g) that contains 80% of carbon and 20% of hydrogen.

7. Phosphorus(III) chloride (one mole = 137.5g) that contains 22.5% of phosphorus and 77.5% of chlorine.

8. Butane (one mole = 58g) that contains 82.8% carbon and 17.2% of hydrogen.

9. A compound X (one mole = 342g) that contains 15.8% of aluminium, 28.1% of sulfur and 56.1% of oxygen. The compound forms a white precipitate when it reacts with barium nitrate solution.

1.3. Balanced equations

As seen in 1.1 the stoichiometric amounts or molar ratios for any reaction can be obtained from a balanced equation.Example: Fe2 O3 (s) + 3CO (g) 2Fe (s) + 3 CO2 (g)

This means that 1 particle + 3 particles 2 particles + 3 particles of Fe2 O3 of CO of Fe of CO

Alternatively this means that there are 3 times as much particles of CO as Fe2 O3 particles; for the reaction to go to completion (=both reactants are completely used up) 3 times as much CO is needed as Fe2 O3.By multiplying each number of particles by 6.02 x 1023, the above balanced equation states the same ratios but expressed in moles:

1 mole reacts with 3 moles to give 2 moles and 3 moles of Fe2 O3 of CO of Fe of CO2

In the above equation the molar ratio of iron oxide to carbon monoxide is 1: 3 i.e. for the reaction to go to completion 3 times more moles of CO2 are needed than Fe2O3

Complete the following table showing other molar ratio’s in the above equation;

Fe2 O3 : Fe :Fe : CO2 :CO : CO2 :

Fe2 O3 : CO2 :

Balancing equations

In getting an equation to balance, the number of moles of each substance is changed, until there are the same number of atoms of each element on each side of the equation:


1.3.1 Deduce chemical equations when all reactants and products are given.1.3.2 Identify the mole ratio of any two species in a chemical equation.1.3.3 Apply the state symbols (s), (l), (g) and (aq).

Unbalanced equation: C3H8 + O2 ---> CO2 + H2O

Balanced equation: C3H8 + 5O2 ---> 3CO2 + 4H2O

Conservation of mass

When applying the molar mass concept to the equation below the Law of Conservation of Mass can be proven: Fe2 O3 (s) + 3CO (g) 2Fe (s) + 3 CO2 (g)

1 x 160 g + 3 x 28g 2 x 56 g + 3 x 44 g 244 g 244 g

the mass before = the mass after the reaction!!!!!Any balanced equation should include state symbols.

When possible you should always apply the state symbols:(s) = solid (l) = liquid (g) = gas (aq) = dissolved in water

1. 4 . Mass and gaseous volume relationships in chemical reactions

Example: How much iron oxide is needed to produce 28g of iron during the reduction of iron oxide by carbon monoxide?

Step 1: write balanced equation

Fe2 O3 (s) + 3CO (g) 2Fe (s) + 3 CO2 (g)

Step 2: convert data in question into moles

data in question = 28 g or iron

number of moles of iron = 28g / 56g = 0.5 mole of iron

Step 3: find answer in equation i.e. use molar ratios

the equation tells us that to get 2 moles of Fe we need 1 mole of Fe2 O3 (= molar ratio)

0.5 mole of iron needs 0.25 moles of Fe2 O3 .

Step 4: convert answer into relevant units

mass = number of moles x molar mass = 0.25g x 160 g = 40 g

To get 28g or iron, 40g of iron oxide is needed to get the reaction to go to completion!!!


1.4.1 Calculate theoretical yields from chemical equations.

Theoretical yield and percentage yield

The quantity of product that you should obtain according to the amounts of reactants available and the molar ratios of reactants is called the theoretical yield.

However, when the reaction is carried out the actual yield is nearly always less than the theoretical yield.

Comparing the actual and theoretical yield by calculating their ratio allows us to determine the efficiency of a reaction; the ratio is called the percentage yield.

actual/experimental yield percentage yield = x 100 theoretical yield

Limiting reagent

If in a chemical reaction the molar ratios of the equation are not followed, usually one of the reactants becomes the limiting reagent whilst all the others are in excess. The limiting reagent controls or limits the amount of product that is formed. At the end of such a reaction in which there is a limiting reagent, besides product, some of the reactants that are in excess will be left.

To determine which reactant is the limiting reagent:

Calculate the number of moles of each reactant; Determine the most simple ratio of the moles From the equation find the required molar ratio so that the reaction can go to completion (i.e. all

reactants turned into products). The reactant which has a lower number of moles than what is required according to the equation is the

limiting reagent.

The number of moles of the limiting reagent is the number of moles used to calculate the theoretical yield of the reaction.

Sample exercise

Consider the Haber process, when 25000 g of nitrogen is reacted with 5000 g of hydrogen to produce ammonia.

N2(g) + 3H2(g) 2NH3(g)

Determine which reactant nitrogen or hydrogen is the limiting reactant? Calculate the theoretical yield of ammonia in grams and the experimental yield as a % of 10390 g of ammonia.

1. Find the actual number of moles of each reactant from the mass of reactant given.

for N2 n = m = 25000 = 892.8 mol


1.4.3 Solve problems involving theoretical, experimental and percentage yield.

1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.

M 28

for H2 n = m = 5000 = 2500 molM 2

2. Find the required number of moles of each reactant from the moles ratio in the reaction.

Required moles of reactant A = Actual moles of reactant B

x Ratio of A

Ratio of B

Required moles of H2 = 892.8 x 3 = 2678.6 mol1

Required moles of N2 = 2500 x 1 = 833.3 mol3

3. Find the limiting reagent

If the: actual moles required moles that reactant is the one that is limiting

So for H2 2500 mol 2678.6 mol so H2 is the limiting reactant

(for N2 892.8 mol 833.3 mol so N2 is not limiting)

4. Find the mass of one of the products formed (also called the theoretical yield)

Use the actual number of moles of the limiting reagent and compare the mole ratios

e.g. to find the number of moles and the mass of NH3 formed

From the mole ratio’s in the equation

H2 : NH3

3 2

if the actual number of moles of the limiting reagent H2 = 2500 mol then using the mole ratio in the equation the actual number of moles of NH3 = 2500 x 2 = 1666.6 mol

3

H2 : NH3

3 2 2500 1666.6

mass NH3 m = n x M

= 1666.6 x 17


= 28332 g

5. Find the Percentage Yield when given the mass of product

e.g. find the percentage yield of 10390 g of ammonia

Experimental Yield % = Experimental Yield (g) x 100Theoretical Yield (g)

= 10390 x 10028332

= 36.7 %

Exercises

1. Consider the reaction: 2 Al + 3 I2 2 AlI3

Determine the limiting reagent when the following are reacted:

(a) 1.2 mol Al and 2.4 mol I2 (b) 1.2 g Al and 2.4 g I2

2. Freon-12 is used as a coolant in refrigerators. It is formed in the following reaction:

3 CCl4 + 2 SbF3 3 CCl2F2 + 2 SbCl3

150 g CCl4 (Mr = 154) is combined with 100 g SbF3 (Mr = 178.8) to give freon-12, CCl2F2

(Mr = 120.9).

(a) Identify the limiting and excess reagents(b) How many grams of freon-12 can be formed?

3. IB Chemistry Exam November 1998

Aspirin, C9H8O4 is made by reacting ethanoic anhydride, C4H6O3 (Mr = 102.1) with 2-hydroxybenzoic acid C7H6O3 (Mr = 138.1) according to the equation:

C4H6O3 + 2 C7H6O3 2 C9H8O4 + H2O

(a) if 15.0 g of 2-hydroxybenzoic acid is reacted with 15.0 g of ethanoic anhydride, determine the limiting reagent in this reaction. 2 marks


(b) Calculate the maximum mass of aspirin that could be obtained in this reaction. 2 marks

(c) If the mass obtained in this experiment was 13.7 g calculate the percentage yield of aspirin. 1 mark

Avogadro’s law to calculate reacting volumes of gases

Gay-Lussac noticed that there is always a very simple ratio between volumes of gases that react together.

For instance,

1 volume of hydrogen + an equal volume of chlorine 2 volumes of hydrogen chloride

1 volume of oxygen + 2 volumes of hydrogen 2 volumes of water vapour

To explain these results, Avogadro proposed that equal volumes of gases (if measured in the same conditions) must have the same number of molecules.

As a result the above volume observation can also be considered as

1 molecule of hydrogen + 1 molecule of chlorine 2 molecules of hydrogen chloride

Avogadro’s Law states that:

Equal volumes of different gases at the same temperature and pressure contain the same number of molecules (or same number of moles)

If the same volume has the same number of particles than the same number of particles must have the same volume so 1 mole at s.t.p. has a volume of 2.24 x 10 -2 m3 mol-1 or 22.4 dm3 mol-1 and this is called the gas molar volume.

At s.t.p. (which is 273K and 1 atm)

volume of 1 mole of a gas = 2.24 x 10-2 m3 mol-1 or 22.4 dm3 mol-1

In other words, if there are x molecules of O2 in 10 ml of oxygen gas, then there are x molecules of N2 in 10 ml of nitrogen gas and there are 2x molecules of CO2 in 20 ml of carbon dioxide gas.

Another way of putting this is that the volume of a gas depends on the amount of moles, n (provided the conditions are the same). Or, if the volume of H2 gas is double the volume of O2 than there are twice as many moles of O2 as H2.

Avogadro’s Law allows equations involving gases to be interpreted directly in terms of volumes.


1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases.1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations.1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.1.4.7 Solve problems using the ideal gas equation, PV = nRT1.4.8 Analyse graphs relating to the ideal gas equation.

For example, in the equation N2 + 3H2 2NH3 for 2.4 dm3 of N2, 7.2 dm3 of H2 is needed to give 4.8 dm3 of NH3.

Question: Cl2 + H2 2HCl

Hydrogen and chlorine react according to the equation above. What will be the result of the reaction of 2.0 moles of H2 and 1.5 moles of Cl2? A. 3.5 mol of HCl B. 1.5mol of HCl and 0.5 mol of H2

C. 2.0 mol of HCl and 0.5 mol of Cl2 D. 3.0mol of HCl and 0.5 mol of H2

Ideal gas equation

What now happens when gases are not at s.t.p.? Is there still a relationship between the number of moles and volume of the gas?

We can use the gas laws and the ideal gas equation which summarises the gas laws.The gas laws describe the effects of changes in temperature, pressure and volume on a fixed mass of an ideal gas.

Simulations showing the gas laws can be found on:

Charles’s Law: http://www.chm.davidson.edu/ChemistryApplets/GasLaws/CharlesLaw.html Boyle’s Law: http://www.chm.davidson.edu/ChemistryApplets/GasLaws/BoylesLaw.html

You can also use the following simulations to study the effects of the above factors.

http://mc2.cchem.berkeley.edu/Java/molecules/index.html http://www2.biglobe.ne.jp/%7Enorimari/science/JavaApp/Mole/e-gas.html

The ideal gas equation

When combining the above gas laws, the ideal gas equation is obtained

pV = constant (Boyle’s Law) V/T = constant (Charles’s Law) P/T = constant If n/p = constant (Avogadro’s Law)

Irrespective of the conditions the gas is in (pressure, temperature, volume), the ratio of pV over nT is always R!!!

The gas constant can have a different value depending on the units it is expressed in. The value you chose depends on the units the raw data is expressed in although we prefer to use the SI units!!!See: http://www.chm.davidson.edu/ChemistryApplets/GasLaws/GasConstant.html for determining the gas constant.See http://resources.schoolscience.co.uk/BAMA/16plus/aerosch3pg1.html


pV = nRT

http://resources.schoolscience.co.uk/BAMA/16plus/aerosch3pg1.html

http://www.chm.davidson.edu/ChemistryApplets/GasLaws/GasConstant.html

http://www2.biglobe.ne.jp/~norimari/science/JavaApp/Mole/e-gas.html

http://mc2.cchem.berkeley.edu/Java/molecules/index.html

http://www.chm.davidson.edu/ChemistryApplets/GasLaws/BoylesLaw.html

http://www.chm.davidson.edu/ChemistryApplets/GasLaws/CharlesLaw.html

When carrying out calculations using the ideal gas equation, we need to ensure we can cancel units!!! Better to use SI units!!

Therefore: R = 8.31 J K-1 mol-1 ( J = Nm) Pressure is in Pa = N m-2, Volume should be in m3 ( 1m3 = 1000 dm3 and 1 dm3 = 1000 cm3 or ml) Temperature should be in K

If the pressure is in atm then:

Either change to Pa (1 atm = 1.013 x 105 Pa) Or use 0.083 L atm K-1 mol-1

The ideal gas equation can be used to solve a variety of problems such as:

calculate the p,V or T values of a gas in a particular condition provided 2 of the 3 quantities are known

calculate the molar mass of a gas:

o we need to know the conditions it is in and the mass of the sample of the gaso we can then calculate n and then using n = mass/Mr calculate the Mr

o we can also use this calculation to identify an unknown gas experimentally !!!

calculate the new value of either p, V or T provided we know what the other 2 have changed into. If we are assuming that when we change 1 or 2 conditions and the mass or amount gas stays the same, then in pV = nRT both n and R are constants and n = pV/T should be the same for both sets of conditions. So

V1 P1 V2 P2 = T1 T2

Worked example

What volume will be occupied by 0.216 g of CO2 at 21 C and at a pressure of 1.32 atm?

1. Calculate the number of moles of the gas: 0.216g/44.0g = 4.91 x 10 -3 mol

2. Express all temperatures in Kelvin: 21 C = 294 K

3. Convert all other units in SI units: 1.32 atm = 1.32 x 1.013 x 105 Pa = 1.34 x 105 Pa

4. Apply pV = nRT

1.34 x 105 x V = 4.91 x 10-3 x 8.314 x 294 V = 8.96 x 10-5 m3 (= 89.6 cm3)

Worked example

A gas occupies 127 cm3 at a pressure of 0.830 atm and at 28C.


(a) What volume will the same amount of gas occupy at 1.00 atm and 0 C

(b) How many moles of gas are present

1. Express all temperatures in Kelvin: 28 C = 301 K and 0 C = 273 K

P1 V1 P2V2

2. Apply: = T1 T2

(no need to change the units to SI units!! when using the above formula)

V1 x P1 x T2 127 x 0.830 x 273 V2 = = = 95.6 cm3

P2 T1 1.00 x 301

3. The number of moles can be calculated by using the ideal gas equation. When using the ideal gas equation, SI units need to be used

n = PV/RT 8.41 x 104 Pa x 1.27 x 10-4 m3 n = = 4.27 x 10 -3 mol 8.314 J K-1 mol-1 x 301 KExercises

1. The greatest volume for one mole of O2 would be expected at

A. 273 K and 1 atm B. 273 K and 2 atm C. 546 K and 1 atm D. 546 K and 2 atm

2. The temperature (in K) is doubled for a sample of gas in a flexible container while the pressure on it is doubled. The final volume of the gas compared with the initial volume will be

A. the same B. twice as large C. four times as large D. half as large

3. When the temperature of a fixed mass and volume of nitrogen is raised from 25C to 50C, the pressure increases because the

A. number of molecules increases.B. molecules expand and occupy a greater volume.C. molecules collide more frequently and energetically with the surface of the containerD. molecules dissociate into separate atoms.

4. For an ideal gas, the variables in one of the pairs below are inversely related to one another (i.e. one increases as the other decreases). Which pair is it?

A. temperature, kinetic energy B. volume, number of moles C. temperature, rate of diffusion D. pressure, volume

5. A 250 cm3 sample of an unknown gas has a mass of 1.42 g at 35 C and 0.85 atmospheres. Which expression gives its molar mass, Mr? (R = 82.05 atm cm3 K-1 mol-1)

1.42 x 82.05x 35 1.42 x 82.05 x 308 1.42 x 250 x 0.85 1.42 x 82.05 x 308


A. B. C. D. 0.25 x 0.85 0.25 x 0.85 82.05 x 308 250 x 0.85

6. A sample of gas has a certain volume at a temperature of 60C. What must the temperature be in order to double the volume if the pressure is kept constant?

A. 120C B. 333C C. 393C D. 666C

7. 125 cm3 of an unknown gas has a mass of 0.725 g at 25C and 0.97 atmospheres. Which expression will give the relative molar mass of the gas? (R = 82.05 cm3 atm K-1 mol-1)

0.725 x 82.05x 25 125 x 0.97 0.725 x 82.05 x 298 0.725 x 82.05 x 298 A. B. C. D. 0.97 x 125 0.725 x 82.05 x 298 0.97 x 0.125 0.97 x 125

For the following MCs you can use your calculator:

8. A 1.25 dm3 sample of air at 25C has a pressure of 85.9 kPa (0.85 atmospheres). What is its volume (in dm3) if its temperature and pressure are changed to 100C and 65.7 kPa (0.65 atm)?

A. 0.76 B. 1.20 C. 1.31 D. 2.05

9. A sample of ethane occupies a volume of 120 cm3 at 30C and 1.0 atm (101kPa). What volume (in cm3) will it occupy at 40C and 1.2 atm (121 kPa)?

A. 103 B. 133 C. 149 D. 192

10. A 225 cm3 sample of a particular gas weighs 0.774 g. What is the approximate molar mass of this gas if the volume was determined at 98.7 kPa (0.974 atm) and 30C?

A. 120 g mol-1 B. 88 g mol-1 C. 86 g mol-1 D. 77 g mol-1

11. A 0.365g sample of a common anaesthetic has a volume of 225 cm3 at 35C and 98.6 kPa (740 mm Hg). What is its molar mass?

A. 42.1 g mol-1 B. 40.8 g mol-1 C. 39.9 g mol-1 D. 4.79 g mol-1

1. 5 . Solutions

solute = a solid, liquid or gas that is dissolved in another substance

solvent = substance, usually a liquid, that will dissolve another substance

solution = a mixture made up from a solute and solvent


1.5.1 Distinguish between the terms solute, solvent, solution and concentration (g dm–3 and mol dm–3).1.5.2 Solve problems involving concentration, amount of solute and volume of solution.

concentration =

amount of solute in moles or grams concentration = volume of solution in dm3

Example: Calculate the volume of H2 SO4 with a concentration of 2 mol/L to neutralise 250 ml of sodium hydroxide with a concentration of 1 mol/L.

Answer:

Step 1: write balanced equation

2 NaOH (aq) + H2 SO4 (aq) Na2 SO4 (aq) + H2 O (l)

Step 2: convert data into moles

use : number of moles concentration = (mol/L) volume (in L)

number of moles of NaOH = 1 mol/L x 0.25 L = 0.25 mole of NaOH

Step 3: find answer in equation in moles

equation tells us: molar ratio NaOH : H2 SO4 = 2 : 1

0.25 mole of NaOH needs to 0.125 moles of H2 SO4.

Step 4: convert answer into relevant unit

volume of H2 SO4 = number of moles/concentration = 0.125 /2 mol/L = 0.0625 L

Tip: Whenever you are given a question in which amounts of all or nearly reactants are given you will very likely need to decide which one is the limiting reagent.

IB Past Paper questions

1. (M07) On complete combustion, a sample of a hydrocarbon compound produces 1.5 mol of carbon dioxide and 2.0 mol of water. What is the molecular formula of this hydrocarbon?

A. C2H2 B. C2H4 C. C3H4 D. C3H8

2. (M07) When excess BaCl2(aq) was added to a sample of Fe (NH4 )2 (SO4 )2 (aq) to determine the amount in moles of sulfate present, 5.02×10−_ mol of BaSO4 was obtained. How many moles of sulfate ions and iron ions were in the sample of Fe (NH4 )2 (SO4 )2?


3. (M07) What volume of 0.500 mol dm-3sulfuric acid solution is required to react completely with 10.0 g

of calcium carbonate according to the equation below? CaCO_ (s) +H2SO4 (aq) →CaSO4 (aq) +H2O (l) +CO2 (g)

A. 100 cm-3 B. 200 cm-3 C. 300 cm-3

D. 400 cm-3

4. (M06) Which of the following quantities has units?

A. Relative atomic mass B. Relative molecular massC. Molar mass D. Mass number

5. (M06) A reaction occurring in the extraction of lead from its ore can be represented by this unbalanced equation: __ PbS + __ O2 → __ PbO + __SO2 When the equation is balanced using the smallest possible whole numbers, what is the coefficient for O

A. 1 B. 2 C. 3 D. 4

6. (M06) The equation for a reaction occurring in the synthesis of methanol is

What is the maximum amount of methanol that can be formed from 2 mol of carbon dioxide and 3 mol of hydrogen?

A. 1 mol B. 2 mol C. 3 mol D. 5 mol

7. (N05) The complete oxidation of propane produces carbon dioxide and water as shown below.

What is the total of the coefficients for the products in the balanced equation for 1 mole of propane?

A. 6 B. 7 C. 12 D. 13

8. (N05) The relative molecular mass (Mr) of a compound is 60. Which formulas are possible for this compound?


A. I and II only B. I and III only C. II and III only D. I, II and

9. (N05) Which aqueous solution contains the most hydrogen ions?

10. (N06) A 4 g sample of sodium hydroxide, NaOH, is dissolved in water and made up to 500 cm3 of aqueous solution. What is the concentration of the resulting solution?

A. 0.1 mol dm-3 B. 0.2 mol dm-3 C. 0.5 mol dm-3 D. 1.0 mol dm-3

11 (N06) Calcium carbonate decomposes on heating as shown below.

CaCO3 → CaO + CO2

When 50 g of calcium carbonate are decomposed, 7 g of calcium oxide are formed. What is the percentage yield of calcium oxide?

A. 7 % B. 25 % C. 50 % D. 75 %

12. (N06) Sodium reacts with water as shown below.

__ Na + __ H2O → __NaOH + __H2

What is the total of all the coefficients when the equation is balanced using the smallest possible whole numbers?

A.3 B. 4 C. 6 D. 7

13. (M05) The equation for the complete combustion of butane is

What is the amount (in mol) of carbon dioxide formed by the complete combustion of three moles of butane?

A. 4 B. 8 C. 12 D. 24


14. (M05) Which solution contains the greatest amount (in mol) of solute?

A. 10.0 cm3 of 0.500 mol dm−3 NaCl B. 20.0cm3 of 0.400 moldm−3 NaCl

C. 30.0cm3 of 0.300 moldm−3 NaCl D. 40.0cm3 of 0.200 moldm−3 NaCl

15. (M05) How many oxygen atoms are present in 0.0500 mol carbon dioxide?

A. 3.01×1022 B. 6.02 ×1022

C. 6.02 ×1023 D. 1.20 ×1024

16. (M05) How many electrons are there in one 24 12 Mg2+ ion?

A. 10 B. 12 C. 14 D. 22

17. (N04) Consider the following equation.

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)

How many moles of CO2 (g) are produced by the complete combustion of 58 g of butane, C4H10 (g) ?

A. 4 B. 8 C. 12 D. 16

18. (N04) 6.0 moles of Fe2O3 (s) reacts with 9.0 moles of carbon in a blast furnace according to the equation below. Fe2O3 (s)+3C(s)→2Fe(s)+3CO(g)

What is the limiting reagent and hence the theoretical yield of iron?

19. (N04) What volume of 0.500 moldm−3 HCl (aq) is required to react completely with 10.0 g of calcium carbonate according to the equation below?

CaCO3 (s) + 2HCl(aq) → CaCl2 (aq) + H2O(l) + CO2 (g)

A. 100 cm3 B. 200 cm3 C. 300 cm3 D. 400 cm3

20. (M04) What amount of NaCl (in moles) is required to prepare 250 cm3 of a 0.200 mol dm-3 solution?

A. 50.0 B. 1.25 C. 0.800 D. 0.0500

21. (M02) A compound that contains only carbon, hydrogen and oxygen has the following percentage by mass:


carbon: 60% hydrogen: 8% oxygen: 32%

What is a possible molecular formula?

A. C5H8O2 . B. C5H4O C. C6HO3 D. C7HO4

22. (M02) Which sample contains the smallest amount of oxygen?

A. 0.3 mol H2SO4 B. 0.6 mol O3 C. 0.7 mol HCOOH D. 0.8 mol H2O

23. (M02) 6.4 g of copper wire is added to 0.10 dm3 of 1.0 mol dm-3 aqueous AgNO3 to form metallic silver and aqueous copper (II) nitrate. When the reaction is complete,

a. excess copper wire remains.b. all the copper wire dissolves and some silver ions are left in solution.c. All the copper wire dissolves and no silver ions are left in solution.d. The mass of metallic silver formed is equal to the mass of copper wire that reacts.

24. (M02) 2.02 g of KNO3 (Mr= 101) is dissolved in sufficient water to prepare 0.500 dm3 of solution. What is the concentration of this solution in mol dm-3?

A. 0.02 B. 0.04 C. 0.10 D. 0.20

25. (M03) What amount of oxygen, O2, (in moles) contains 1.8 x 1022 molecules?

A. 0.0030 B. 0.030 C. 0.30 D. 3.0

26. (M03) 3.0 dm3 of sulfur dioxide are reacted with 2.0 dm3 of oxygen according to the equation: 2SO2 (g) + O2 (g) 2SO3 (g)

What volume of sulfur trioxide (in dm3) is formed? (Assume the reaction goes to completion and all gases are measured at the same temperature and pressure. A. 5.0 B. 4.0 C. 3.0 D. 2.0

27. (M03) _ C2H2 (g) + _ O2 (g) _ CO2 (g) + _ H2O (g)

When the equation is balanced, what is the coefficient of oxygen?

A. 2 B. 3 C. 4 D. 5

28. (N03) Copper can react with nitric acid as follows

3Cu + _ HNO3 _ Cu(NO3)2 + _ H2O + _NO

What is the coefficient for HNO3 when the equation is balanced?


A. 4 B. 6 C. 8 D. 10

29. (N03) A hydrocarbon contains 90% by mass of carbon. What is its empirical formula?

A. CH2 B. C3H4 C. C7H10 D. C9H10

30. (N03) Lithium hydroxide reacts with carbon dioxide as follows

2LiOH + CO2 Li2CO3 + H2O

What mass (in grams) of lithium hydroxide is needed to react with 11g of carbon dioxide?

A. 6 B. 12 C. 24 D. 48

31. (M04) How many hydrogen atoms are contained in one mole of ethanol, C2H5OH?

A. 5 B. 6 C. 1.0 x 1023 D. 3.6 x 1024

32. (M04) The percentage by mass of the elements in a compound is C = 72% H = 12% O = 16%

What is the mole ratio C : H in the empirical formula of its compound?

A. 1 : 1 B. 1 : 2 C. 1 : 6 D. 6 : 1

33. (M04) What is the coefficient for O2 (g) when the equation below is balanced?

_ C3H8 (g) + _ O2 (g) _ CO2 (g) + _ H2O (g)

A. 2 B. 3 C. 5 D. 7

34. According to the equation 2SO2 (g) + O2 (g) 2SO3 (g)

what volume of air (20% O2) is required to react with 10 dm3 of SO2?

A. 2 dm3 B. 5 dm3 C. 10 dm3 D. 25 dm3

35. (M03) Which of the following compounds has the greatest empirical formula mass? A. C6H6 B. C4H10 C. C5H10 D. C2H6

36. CaCO3 (s) CaO(s) + CO2 (g)

When heated, CaCO3 (Mr = 100) decomposes as shown above. When 20 g of impure CaCO3 is heated, 0.15 moles of CO2 are obtained. What is the % purity of the CaCO3 ?


A. 15 B. 25 C. 55 D. 75

Paper 2

1. (M07) (a) Propane and oxygen react according to the following equation. C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(g) Calculate the volume of carbon dioxide and water vapour produced and the volume of oxygen remaining, when 20.0 dm3 of propane reacts with 120.0 dm3 of oxygen. All gas volumes are measured at the same temperature and pressure. [3]

(b) State and explain what would happen to the pressure of a given mass of gas when its absolute temperature and volume are both doubled. [3]

2. (N05) An organic compound, A, containing only the elements carbon, hydrogen and oxygen was analysed.

(a) A was found to contain 54.5 % C and 9.1 % H by mass, the remainder being oxygen. Determine the empirical formula of the compound. [3]

(b) A 0.230 g sample of A, when vaporized, had a volume of 0.0785 dm3 and 102 kPa at 95 C °. Determine the relative molecular mass of A. [3] (c) Determine the molecular formula of A using your answers from parts (a) and (b). [1]

3. (M06) The reaction below represents the reduction of iron ore to produce iron.

A mixture of 30 kg of and 5.0 kg of C was heated until no further reaction occurred. Calculate the maximum mass of iron that can be obtained from these masses of reactants. [5]

4. (N06) An organic compound A contains 62.0 % by mass of carbon, 24.1 % by mass of nitrogen, the remainder being hydrogen.

(i) Determine the percentage by mass of hydrogen and the empirical formula of A. [3]

(ii) Define the term relative molecular mass. [2] (iii) The relative molecular mass of A is 116. Determine the molecular formula of A. [1]

5. (M05) The percentage composition by mass of a hydrocarbon is C = 85.6 % and H = 14.4 %. (a) Calculate the empirical formula of the hydrocarbon. [2]

(b) A 1.00g sample of the hydrocarbon at a temperature of 273K and a pressure of 1.01×105 Pa (1.00 atm) has a volume of 0.399 dm3.Topic1notes 12.5 hours Page 24 of 31

(i) Calculate the molar mass of the hydrocarbon. [2] (ii) Deduce the molecular formula of the hydrocarbon. [1]

6. (M03) (a) The relative molecular mass of aluminium chloride is 267 and its composition by mass is 20.3% Al and 79.7% chlorine. Determine the empirical formula and molecular formulas of aluminium chloride. [4]

(b) Sodium reacts with water as follows. 2Na(s) + 2H2O (l) 2NaOH (aq) + H2 (g)

1.15 g of sodium is allowed to react completely with water. The resulting solution is diluted to 250 cm3. Calculate the concentration, in mol dm-3, of the resulting sodium hydroxide solution. [3]

7. (N03) (a) Aqueous XO4

3- ions form a precipitate with aqueous silver ions, Ag+. Write a balanced equation for the reaction, including state symbols. [2]

(b) When 41.18 cm3 of a solution of aqueous silver ions with a concentration of 0.2040 mol dm -3 is added to a solution of XO4

3- ions, 1.172 g of the precipitate is formed.

(i) Calculate the amount (in moles) of Ag+ ions used in the reaction. [1]

(ii) Calculate the amount (in moles) of the precipitate formed. [1]

(iii) Calculate the molar mass of the precipitate. [2]

(iv) Determine the relative atomic mass of X and identify the element. [2]

8. (M02) A student was asked to make some copper (II) sulfate-5-water (CuSO4.5H2O) by reacting copper (II) oxide (CuO) with sulfuric acid.

(a) Calculate the molar mass of copper (II) sulfate-5-water. [1]

(b) Calculate the amount (in mol) of copper (II) sulfate-5-water in a 10.0 g sample. [1]

(c) Calculate the mass of copper (II) oxide needed to make this 10.0 g sample. [1]

9. (M03) (a) Write an equation for the reaction between hydrochloric acid and calcium carbonate. [2](b) Determine the volume of 1.50 mol dm-3 hydrochloric acid that would react with exactly 1.25 g of

calcium carbonate. [3] (c) Calculate the volume of carbon dioxide, measured at rtp, which would be produced when 1.25 g

of calcium carbonate reacts completely with the hydrochloric acid. [2]


10. (M01) (a) An anti-cancer drug called Cisplatin has the following percentage composition by mass: Pt = 65.01% Cl = 23.63% N = 9.340 % H = 2.020 %

Calculate the empirical formula of Cisplatin. [3]

(b) The molecular and empirical formulas of Cisplatin are the same. Analysis of the molecule shows platinum to be the central atom, being bonded to four separate atoms; the hydrogen is bonded to nitrogen. Draw the representation of the molecule. [1]

(c) 16.20 x 10-3 dm3 of 0.1020 mol dm-3 aqueous AgNO3 is added to 14.80 x 10-3 dm3 of 0.125 mol dm-3 aqueous NaCl. Calculate the maximum mass (g) of AgCl which could be obtained from this reaction. [4]

11. (N01) Indigo is a blue dye which contains only carbon, nitrogen, hydrogen and oxygen.

(a) 2.036 g of indigo was completely oxidized to produce 5.470 g of carbon dioxide and 0.697 g of water. Calculate (i) the percentage of mass of carbon in indigo. [2] (ii) the percentage of mass of hydrogen in indigo. [2]

(b) If the percentage by mass of nitrogen in the indigo sample is 10.75%, determine the empirical formula of indigo. [3]

(c) If the molar mass is approximately 260 g mol-1, determine the molecular formula of indigo. [2]

12. (N02) A balloon, which can hold a maximum of 1100 cm3 of air before bursting, contains 955 cm3 of air at 5ºC . Determine whether the balloon will burst if the temperature is increased to 25 C. Assume that the pressure of the gas in the balloon remains constant. [3]

13. (N02) An element X reacts with oxygen to form the oxide X2O3 .

(a) Write a balanced equation for the reaction. [1]

(b) If 2.199 g of the oxide was obtained from 1.239 g of X, calculate the relative atomic mass of X and identify the element. [5]

(c) Nitrogen also forms an oxide on reaction with oxygen. This oxide contains 25.9 % of nitrogen and 74.1 % of oxygen by mass. Calculate the empirical formula of this second oxide. [3]

14. (M02) A student is asked to prepare some copper(II) nitrate by reacting nitric acid with copper(II) oxide. (a) Write a balanced equation for this reaction. [1] (b) The student carries out this reaction by adding 0.0345 mol of copper(II) oxide to 36.0 of cm3

1.15 mol dm-3 nitric acid solution. Calculate the amount (in mol) of nitric acid. [1]


(c) Use the information in (a) and (b) to identify the limiting reagent and determine the amount (in mol) of copper(II) nitrate formed. [2]

15. The value of X in NaAl(SO4)2.XH2O can be found by determining the number of moles of sulphate in the compound quantitatively. A 5.218g sample was dissolved and excess BaCl2 was added. The precipitate of BaSO4 was separated, dried and found to weigh 5.315g. (use integer mass numbers).

(a) Calculate the number of moles of BaSO4 in the 5.315g sample.(b) How many moles of sulphate are there in the 5.218g of NaAl(SO4)2.XH2O?(c) Calculate the number of moles of Al in the 5.218g of NaAl(SO4)2.XH2O.(d) Determine the total mass of sulphate, sodium and aluminium that are present in the 5.218g sample

of NaAl(SO4)2.XH2O(e) Determine the number of moles of water in the 5.218g sample of NaAl(SO4)2.XH2O(f) Determine the value of X in NaAl(SO4)2.XH2O

Mark scheme

PAPER 1

1 11 21 31 412 12 22 32 423 13 23 33 434 14 24 34 445 15 25 35 456 16 26 36 467 17 27 37 478 18 28 38 489 19 29 39 49


10 20 30 40 50

PAPER 2

1. (M07)

2. (N05)

1. (M06)

2. (N06)


4.(M01)


5. (N01)


6. (N02)

7. (N02)

8.(M02)


Documents

Topic 1 Quantitative Chemistry