MASS RELATIONSHIPS IN CHEMICAL REACTIONScheminnerweb.ukzn.ac.za/Files/Stoichiometry.pdf · MASS RELATIONSHIPS IN CHEMICAL REACTIONS 1. The mole, Avogadro’s number and molar mass

Chemistry for Engineers 1A (CHEM 181) – UKZN – School of Chemistry , Howard College Campus, compiled by A.Bissessur 2011

1

MASS RELATIONSHIPS IN CHEMICAL REACTIONS

1. The mole, Avogadro’s number and molar

mass of an element

2. Molecular mass (molecular weight)

3. Percent composition of compounds

4. Empirical and Molecular formulas

5. Stoichiometry

6. Limiting Reagents

7. Reaction Yields

8. Combustion reactions


2

THE MOLE (unit mol, symbol n)

It is the amount of a substance that contains as many elementary entities (atoms, molecules or other particles) as there are atoms in exactly 12 g of the carbon-12 isotope.

Consider: 1 mol = 6.022 x 1023 and 1 mol of C = 12 g of C

therefore the actual number of atoms in 12 g of the carbon-12 isotope = 6.022 x 1023 atoms called the Avogadro’s Number

1 mol of C = 12 g = 6.022 x 1023 atoms

Converting between moles and number of atoms

1 mol atoms 6.022 x 1023 atoms

or 6.022 x 1023 atoms 1 mol atoms

Example 1.

Given: 3.5 mol He Find: He atoms

Solution: 3.5 mol He x 6.022 x 1023 He atoms 1 mol He atoms = 2.1 x 1024 He atoms

Example 2.

Given: 1.1 x 1022 Ag atoms Find: mols Ag

Solution: 1.1 x 1022 Ag atoms x 1 mol Ag atoms 6.022 x 1023 Ag atoms

= 1.8 x 10-2 mols Ag


3

CONVERTING BETWEEN GRAMS AND MOLES OF AN ELEMENT

The mass of 1 mol of atoms of an element is called the molar mass

The value of an element’s molar mass in grams per mole is numerically equal to the element’s atomic mass in atomic mass units (amu)

For example Copper has a atomic mass of 63.55 amu, therefore 1 mol of copper atoms has a mass of 63.55 g and the molar mass of copper = 63.55 g/mol or g mol-1.

32.07 g of sulfur = 1 mol sulfur = 6.022 x 1023 S atoms

12.01 g of carbon = 1 mol carbon = 6.022 x 1023 C atoms

6.94 g of lithium = 1 mol lithium = 6.022 x 1023 Li atoms

Therefore the molar mass of any element becomes a conversion factor between grams of that element and moles of that element:

Example for carbon:

12.01 g = 1 mol C = 12.01 g C1 mol C

or 1 mol C

12.01 g C

Example 1.

Given: 0.58 g of C Find: mols of C

Solution: 0.58 g x 1 mol C

12.01 g C = 4.8 x 10-2 mols

Mass of element (g) 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚→

mols of element (n) n x 𝑁𝑁𝐴𝐴

→ Number of atoms (N)

Mass of element (g) n x molar mass

← mols of element (n)

N/𝑁𝑁𝐴𝐴←

Number of atoms (N)

N = no of atoms, NA = Avogadro’s number, n = no. of moles


4

COUNTING MOLECULES BY THE GRAM

For elements, molar mass is the mass of 1 mol of atoms of that element.

For compounds, the molar mass is the mass of 1 mol of molecules or formula units of that compound (also known as molecular mass or molecular weight)

If the atomic masses of component atoms, the mass of the molecule can be calculated.

For example the molar mass /molecular mass/molecular weight or formula mass of CO2 is:

Molecular mass = 1(atomic mass of C) + 2(atomic mass of O)

= 1(12.01 amu) + 2(16.00 amu)

= 44.01 amu

The molar mass of CO2 = 44.01 g/mol or g mol-1

CONVERTING BETWEEN GRAMS AND MOLES OF A COMPOUND

Given: 22.5 g of CO2 Find: mols CO2

Solution: 22.5 g x 1 mol CO2

44.01 g = 0.511 mol CO2

CONVERTING BETWEEN GRAMS OF A COMPOUND AND NUMBER OF MOLECULES

Example 1.

Given: 22.5 g of CO2 Find: number of CO2 molecules

Solution: 22.5 g x 1 mol CO2

44.01 g x 6.022 x 1023 CO2 molecules

mol CO2 = 3.08 x 1023 CO2 molecules


5

Example 2

Given: 4.78 x 1023 NO2 molecules Find: grams NO2

Solution:

Molar mass of NO2 = 14.01 + 2(16.00) = 46.01 g mol-1

4.78 x 1023 NO2 molecules X 1 mol NO2

6.022 x 1023 NO2 molecules X

46.01 g NO21 mol NO2

= 36.5 g NO2

Practice Exercise

1. Calculate the molar mass of the following compounds:

(a) N2 (b) CO (c) MgCl2 (d) CaCl2. 6H2O (e) C6H5OH

(f) (NH4)SO4.FeSO4.6H2O (g) CH3CH2CH2CH2CH2COOH

Percent Composition of Compounds

It is the percentage mass of each element in a compound.

General formula:

% composition of element X = n x molar mass of element X molar mass of compound

x 100%

Where n = number of moles of the element in 1 mole of the compound


6

Example 1.

Calculate the mass % of H and O in H2O2.

Solution:

1 mol of H2O2 = 34.02 g contains 2 mols of H and 2 mols of O

Molar masses (in g mol-1): H2O2 = 34.02, H = 1.008, O = 16.00

% H = 2 x 1.008 H g/mol34.02 H2O2 g/mol

x 100 = 5.926 %

% O = 2 x 16.00 O g/mol34.02 H2O2 g/mol

x 100 = 94.06 %

Calculation of the percent composition of H and O in the Empirical formula of H2O2 which is HO

% H =1 x 1.008 H g/mol17.01 HO g/mol

x 100 = 5.926 %

% O = 1 x 16.00 O g/mol17.01 HO g/mol

x 100 = 94.06 %

Both molecular formula and empirical formula gives the same percent composition.

Practice Exercise

1. Calculate the percent composition of Cl in CCl2F2.

2. Calculate the percent composition of each element in C12H23O6Br

3. Calculate the percent composition of K and Mn in KMnO4

4. Calculate the percent composition of H

.

2O in CaCl2⋅ 2H2O


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CALCULATING EMPIRICAL AND MOLECULAR FORMULAS FROM REACTION DATA

(a) Calculating Empirical Formulas

Example 1.

Calculate the empirical formula of a compound that is made up of 69.58% Ba, 6.090 % C and 24.32 % O only.

Solution

Mass % →mass of element →moles of each element → mole ratio → empirical formula

Assume 100g of the compound

Therefore: Ba = 69.58 g, C = 6.090 g , O = 24.32 g

Ba C O

Mass (g) 69.58 6.090 24.32

Molar mass (g mol-1

Empirical formula = BaCO

) 137.3 12.01 16.00

Moles (mols) 69.58137.3

6.09012.01

24.3216.00

0.5068 0.5071 1.520

Divide by the smallest mols

0.50680.5068

0.50710.5068

1.520

0.5068

1 1 2.999

Round off to the nearest whole number

1 1 3

3


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Practice Exercise 1.

A compound containing nitrogen and oxygen is decomposed in the laboratory

and produces 24.5 g of nitrogen and 70.0 g of oxygen. Determine the empirical

formula of the compound.

Solution: N O

Mass (g) 24.5 70.0

Molar mass (g mol-1) 14.01 16.00

Moles (mols) 24.5

14.01

70.016.00

1.75 4.38

Divide by the smallest mols

1.751.75

4.381.75

1 2.5

Multiply by a factor# to convert to a whole number

N1O2.5 x 2 = N2O5

# See notes below


A laboratory analysis of aspirin determined the following mass percent

composition: C = 60.00 %, H = 4.48 % and O = 35.53 % only.

Find the empirical formula.

Answer: C9H8O4


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If subscripts are not whole numbers, multiply all the subscripts by a small whole number to get whole number subscripts

Fractional subscript Multiply by this number to get

Whole-number subscripts

__.10 10

__.20 5

__. 25 4

__.33 3

__.50 2

__.66 3

__.75 4


A sample was found to contain 13.42 g of C, 2.25 g of H and 17.88 g of O only.

Determine the empirical formula for this compound.


A sample has the following percentage compositions:

40% C, 6.71% H and 53.29% O

What is the empirical formula for this compound?

Practice Exercise 5

A 3.24 g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide?


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CALCULATING MOLECULAR FORMULAS FOR COMPOUNDS

To calculate the molecular formula, the following must be known:

(i) molar mass of the compound

(ii) empirical formula of the compound

Note: The formula calculated from percent composition is always the empirical formula

Molecular formula = Empirical formula × n, where n = 1,2,3……

Example:

Find the molecular formula for fructose (a sugar formed in fruit) from its empirical formula CH2O and its molar mass, 180.2 g mol-1.

Solution:

Molar mass is a whole-number multiple of the empirical formula mass, the sum of the masses of all atoms in the empirical formula.

Therefore molar mass = Empirical formula × n

and n = Molar mass

Empirical formula molar mass

Empirical formula molar mass for fructose = 1(12.01) + 2(1.01) + 16 = 30.03 g mol-1

Therefore, n = 180.2 g/mol30.03 g/mol

= 6, hence molecular formula = CH2O × 6= C6H12O6

Practice Exercise 1:

Given that the empirical formula of a compound is CH and the molar

mass is 104 g mol-1

, determine the molecular formula.


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Solution:

Empirical formula mass = 12.01 g mol-1 + 1.01 g mol-1 = 13.01 g mol-1

Divide the molar mass of the compound by the empirical formula mass:

No of CH units = 104 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚−1

13.01 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚−1 = 8.00

Therefore molecular formula = CH × 8 = C8H8


Naphthalene , a compound containing carbon, and hydrogen only, is often used in moth balls. Its empirical formula is C5H4 and its molar mass is 128.16 g mol-1. Find its molecular formula.

Solution: C10H

8

STOICHIOMETRY Greek: Stoicheon = element metron = element measuring

Stoichiometry is the science of measuring the quantitative proportions or mass ratios in which chemical elements stand to one another

Molar Ratios:

xA + yB → aA + zD Stoichiometric relationship:

nAx

= nBy

= n𝐶𝐶a

= n𝐷𝐷z


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Example 1:

Consider the following balanced equation and write the stoichiometric relationship for all reactants and products

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O (l)

Solution:

nC2H6(g)

2 =

nO2(g)

7 =

nCO2(g)

4 =

nH2O(g)

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1. Mole to Mole conversions: mol → mol

3 H2(g) + N2(g) → 2 NH3(g)

3 H2 molecules ≡ 1 N2 molecule ≡ 2 NH3 molecules

3 mols H2 ≡ 1 mol N2 ≡ 2 mol NH3

Example 1.

If 2.0 mol of N2(g) reacts with sufficient H2(g), how many mols of NH3(g) will be produced?

Solution:

nN2

1 =

nNH32

⇒ 2.0 mols

1 =

nNH32

nNH3 = 2.0 mols x 2 = 4.0 mols

Example 2.

If 4.25 mols of H2(g) reacted with sufficient N2(g), calculate the number of moles of NH3

(g) that would be produced.


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2. Mole to mass conversions: mol → grams

Example 1:

How many grams of oxygen are produced when 1.50 mols of KClO3

2 KClO

(s) are decomposed according to the balanced equation?

3(s) 2 KCl(s) + 3 O2(g)

nKClO32

= nO2

3 ⇒

1.5 mols2

= nO2

3

Therefore nO2 = 1.50 mols × 32 = 2.25 mols

nO2 = mass of O2

Molar mass of O2 ⇒ mass of O2 = 2.25 mols × 32.00 g mol

3. Mass to mole conversions

-1

= 72.0 g

2 KClO3(s) 2 KCl(s) + 3 O2(g)

Example:

If 80.0 g of O2(g) was produced in the above reaction, calculate the number of moles of KClO3 decomposed.

Solution:

Stoichiometry from balanced equation is:

nKClO32

= nO2

3 ⇒ nKClO3 = nO2 ×

23 = 80 g

32.00 g/mol ×

23

= 1.67 mols of KClO3 decomposed

∆

∆


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Practice Exercise:

Consider the following equation: 2H2(g) + O2(g) → 2H2O

(a) How many grams of H2O are produced when 2.50 moles of O2(g) is reacted?

(b) If 3.00 moles of H2O is produced, calculate the mass of O2(g) that was used.

(c) How many grams of H2(g) must be used, given the data in (b) above?

4. Mass to mass conversions

Example 1

How many grams of Cl2(g) can be liberated from the decomposition of 64.0 g of AuCl3 in the following reaction:

2 AuCl3(aq) → 2 Au(s) + 3 Cl2

Solution:

From stoichiometry

nAuCl32

= nAu

2 =

nCl23

nAuCl32

= nCl2

3

nCl2 = nAuCl3

2 x 3 =

64.0 g303.32 g/mol

× 32 = 0.211 mols

mass of Cl2(g) = 0.211 mols × 70.9 g mol-1

= 22.40 g


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Practice Exercises

1. Calculate the mass of AuCl3 that can be produced from 100 g of Cl2 in the following reaction:

2 Au(s) + 3 Cl2 → 2 AuCl3(aq)

2. Calculate the mass of AgCl(s) that can be produced by reacting 200.0 g of AlCl3 and sufficient AgNO3, using the following reaction:

3 AgNO3(aq) + AlCl3(aq) → 3 AgCl(s) + Al(NO3)3 (aq)

3. Using the following reaction:

2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)

Calculate the mass of PbI2(s) by reacting 30.0 g of KI with excess Pb(NO3)2.

4. How many grams of Na(s) are required to react completely with 75.0 g of Cl2(g) using the following equation:

2 Na(s) + Cl2(g) → NaCl (unbalanced)

5. A component of acid rain is sulfuric acid which forms when SO2(g), a pollutant reacts with oxygen and rain water according to the following reaction:

2 SO2(g) + O2(g) + H2O(ℓ) → H2SO4(aq)

Assuming that there is plenty of O2(g) and H2O(ℓ), how much H2SO4 in kilograms forms from 2.6 x 103 kg of SO2(g)?


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5. Limiting Reagents

The limiting reagent (or reactant) is the reactant that is completely consumed in a chemical reaction.

The maximum amount of product formed depends on how much of this (limiting) reactant was originally present.

Excess reagents are present in quantities greater than necessary to react with the quantity of the limiting reagent.

Theoretical yield - the amount of product that can be made in a chemical reaction based on the amount of limiting reagent.

Actual or (experimental) yield – the amount of product actually produced by a chemical reaction.

Percentage Yield = Actual Yield

Theoretical Yield x 100

Consider a recipe to bake pancakes:

1 cup flour + 2 eggs + ½ tsp baking powder → 5 pancakes

Suppose we have:

3 cups flour + 10 eggs + 4 tsp baking powder → ? pancakes

We can make:

3 cups flour → 15 pancakes

10 eggs → 25 pancakes

4 tsp baking powder → 40 pancakes

Flour is the limiting reagent as it produces the least amount of pancakes


17


Consider the following reaction:

Ti(s) + 2Cl2(g) → TiCl4(s)

If we begin with 1.8 mol of Ti and 3.2 mol of Cl2, what is the limiting reagent and calculate the theoretical yield of TiCl4 in moles?

Solution:

Given: 1.8 mol Ti Find: limiting reagent

3.2 mol Cl2 Theoretical yield

Stoichiometry:

nTiCl41

= nCl2

2 =

3.2 mols2

= 1.6 mols

nTiCl4

1 =

nTi1

= 1.8 mols

Since smaller amount of mols of TiCl4 is produced from Cl2,

Therefore Cl2 is the limiting reagent while Ti is the excess reagent.

Therefore theoretical yield of TiCl4(s) = 1.6 mols



2 Al(s) + 3Cl2(g) → 2 AlCl3(s)

If we begin with 0.552 mol of aluminium and 0.887 mol of chlorine, what is the limiting reagent and the theoretical yield?


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6. Limiting Reagent, Theoretical Yield, and Percent Yield from Initial masses of Reactants


2 Na(s) + Cl2(g) → 2 NaCl(s)

If we begin with 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant and theoretical yield?

nNa = 53.2g

22.99g/mol = 2.31 mols, nCl2 =

65.8g70.90g/mol

= 0.928 mols

nNaCl

2 =

nNa

2 = 2.31 mols

nNaCl

2 =

nCl21

⇒ nNaCl = 0.928 mols x 2 = 1.856 mols

Therefore the limiting reagent is Cl2

Theoretical yield (calculated from Cl2)

Mass of NaCl = 1.856 mols x 58.44 g mol-1 = 108 g NaCl

Suppose when the synthesis was carried out, the actual yield of NaCl was found to be 86.4 g. What is the percent yield?

Percentage Yield = Actual Yield

Theoretical Yield x 100

= 86.4 g108 g

x 100

= 80.0 %


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Practice Exercise

1. Ammonia can be synthesized by the Haber Process according to the following reaction:

3 H2(g) + N2(g) → 2 NH3(g)

(a) What is the maximum amount of ammonia in grams that can be synthesized from 25.2g of N2(g) and 8.42g of H2(g)?

(b) What is the maximum amount of ammonia in grams that can be synthesized from 5.22g of H2(g) and 31.5 of N2(g)?

2. Consider the following reaction:

Cu2O(s) + C(s) → 2 Cu(s) + CO2(g)

When 11.5 g of C are allowed to react with 114.5 g of Cu2

O(s), 87.4 g of Cu are obtained. Find the limiting reagent , theoretical yield and percent yield.


20

COMBUSTION REACTIONS A combustion reaction is one in which the elements in a compound react with

molecular oxygen to form the oxides of those elements. For example C in a

carbon-containing compound will be converted to CO2 and if there is

hydrogen it will be converted to H2O

Notice that by accurately measuring the mass of CO2 obtained by combustion

of the carbon-containing compound, the mass of carbon in the original sample

can be calculated.

Similarly, by measuring the mass of H2O formed in the reaction, the mass of

hydrogen in the original sample can be calculated. These calculations assume

that all the carbon in the sample is captured in the CO2 and that all the

hydrogen is captured in the H2O

EXAMPLE 1: Combustion reaction involving C,H and O only

Vitamin C is a compound that contains the elements C. H and O. Complete

combustion of a sample of mass 0.2000 g of vitamin C produced 0.2998 g of

CO2 and 0.08185 g of H2O. Determine the empirical formula of vitamin C.

Molar masses (in g mol-1) :

CO2 = 44.01 H2O = 18.02 ; C = 12.01 H = 1.008 O = 16.00

CxHyOz +O2 → x CO2 2y + H2

All the C is converted to CO

O

2 and all the H is converted to H2O.


21

Solution:

Step1: Determine the mass of C in CO2

12.01 g mol −1

44.01 g mol −1 x 0.2998 g CO2 = 0.08181 g of C

Step 2: Calculate the mass of water hydrogen in water

2 x 1.008 g mol −1

18.02 g mol −1 x 0.08185 g H2

÷ by smallest number of moles

O = 0.009157g of H

(note: there are 2 mols of hydrogen in water)

Step 3: Calculate the mass of O which is obtained by difference

Mass of O = Mass of sample – (mass of C + mass of H)

= 0.2000 g – (0.08181 g + 0.00915 g)

= 0.1090 g of O

Step 4: Convert to moles ( massAtomic mass

)

C H O

Moles 0.006812 0.009084 0.006814

0.0068120.006812

0.0090840.006812

0.0068140.006812

1 1.33 1

convert to whole numbers by multiplying by 3

1 x 3 1.33 x 3 1 x 3

3 4 3

Therefore the empirical formula of vitamin C is C3H4O3


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EXAMPLE 2. Combustion reaction involving C,H,O and N only

The compound caffeine contains the elements C, H, N and O. Combustion

analysis of a 1.500 g sample of caffeine produces 2.737g of CO2 and 0.6814 g

of H2O. A separate further analysis of another sample. of mass 2.500 g of

caffeine produces 0.8677 g of NH3.

Determine the empirical formula of caffeine.

Molar masses(in g mol-1) : H = 1.008 C = 12.01 ; N = 14.01 ; O = 16.00

CO2 = 44.01 ; H2O = 18.02 ; NH3

= 17.04.

Important features of this problem are that:

1. You are analysing for 4 elements – C, H, N and O.

2. The analysis is performed on two samples (for example sample 1 and 2)

which have different masses.

NOTE:

In the one sample you analyse for carbon and hydrogen.

In the other you analyse for nitrogen. Remember that the oxygen is always

obtained by difference.


23

Solution:

In sample 1 - 1.500g of caffeine

Step1: Determine the mass of C in CO2

12.01 g mol −1

44.01 g mol −1 x 2.737 g CO2 = 0.7469 g of C

Step 2: Calculate the mass of water hydrogen in water

2 x 1.008 g mol −1

18.02 g mol −1 x 0.06814 g H2O = 0.07623 g of H

In sample 2 - 2.500 g of caffeine

Step 3. Determine the mass of N in NH3

14.01 g mol −1

17.04 g mol −1 x 0.8677 g NH3

= 0.7134 g of N

Step 4: Calculate % N in sample 2 = % N in sample 1

Note: The % composition of a pure compound is constant. If you know the percentage of N in

sample 1 then the percentage of nitrogen in sample 2 is exactly the same.

Therefore % N in sample 2 = 0.7134 g2.500 g

x 100 = 28.54 %

Step 5. You must determine the masses of all the elements in the compound in a common sample therefore find the mass of N in sample 1

i.e. 28.54 g100 g

X 1.500 g = 0.4280 g mass of N


24

Step 6. The mass of O in sample 1 is obtained by difference

Mass of O = Mass of sample 1 – (mass of C + mass of H + mass of N)

= 1.500 g – (0.7469 g + 0.07623 g + 0.4280 g)

= 0.2489 g

Determine the empirical formula

C H N O

Mass (g) 0.7469 0.07623 0.4280 0.2489g

Atomic mass (g mol-1) 12.01 1.008 14.01 16.00

moles: 0.06219 0.07563 0.03057 0.01555

Divide by the smallest number of mol = 0.01555

4 4.9 2 1

~5

∴ Empirical formula is = C4H5N2O


25

PRACTICE EXERCISES

1. An 0.1888g sample of a hydrocarbon produces 0.6260g of CO2 and

0.1602g of H2O in combustion analysis. Its molecular weight is found to

be 106 amu.

For this hydrocarbon determine

(a) its mass percent composition Ans. 90.47% C. 9.50% H

(b) its empirical formula Ans. C4H5

(c ) its molecular formula Ans. C8H10

2. Dimethylhydrazine is a carbon–hydrogen-nitrogen containing

compound. Combustion analysis of a 0.312g sample of the compound

produces 0.458 g of CO2. From a separate 0.525 g sample the nitrogen

content is converted to 0.244 g N2.

(a) What is the empirical formula of dimethylhydrazine? Ans. CH4N

(b) If the molecular weight was found to be 60.02 what is the

molecular formula of the compound? Ans. C2H8N2

3. A 1.35 g sample of a substance containing C. H. N and O is burned in air

to produce 0.810 g of H2O and 1.32 g of CO2. In a separate analysis of

the same substance all of the N in a sample of mass 0.735 g is converted

to 0.284g NH3

Ans. CH

. Determine the empirical formula of the substance.

3NO

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MASS RELATIONSHIPS IN CHEMICAL REACTIONScheminnerweb.ukzn.ac.za/Files/Stoichiometry.pdf · MASS RELATIONSHIPS IN CHEMICAL REACTIONS 1. The mole, Avogadro’s number and molar mass