Gravimetric analysisFour example problems

1. Determining salt in food2. Determining sulfate and sulfur content in fertiliser3. Determining the empirical formula of a hydrated

molecule4. Determining the empirical formula of an alcohol

Question 1: Determining salt in foodA 6.67 g sample of potato chips was crushed, blended with water and filtered. Excess silver nitrate was added to the filtrate. The dried precipitate had a mass of 0.245 g. Calculate the amount of sodium chloride that was in the potato chips (assume all chloride is in the form of the sodium salt). Give your answer in mg of NaCl per 100 g of chips.

Where do we start?• Write out the relevant chemical reaction:

NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)

• Identify what else we know:The final mass of AgCl is 0.245 g (3 significant figures) taken from 6.67 g (3 significant figures) of chips

• Think about where you want to get to:Calculate how many mg of NaCl are present in 100 g of potato chips

Calculating the moles of AgCl•NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)

m(AgCl) = 0.245 gM(AgCl) = 107.9 gmol-1 + 35.5 gmol-1

= 143.4 gmol-1

n(AgCl) = 0.245 g/143.4 gmol-1

= 0.001709 mol

Calculating the mass of NaCl• NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)

n(AgCl) = 0.001709 molFrom the chemical equation we know that the moles of NaCl to the moles of AgCl is 1:1, therefore:

n(NaCl) = 0.001709 molM(NaCl) = 23.0 gmol-1 + 35.5 gmol-1 = 58.5 gmol-1

m(NaCl) = 0.001709 mol × 58.5 gmol-1 = 0.09998 g

Calculating the amount of NaCl in mg per 100 g

There are mg of NaCl per 100 g of chips.

What does this really mean?• If you look at the nutrition panel of any food, you will find the mass of

sodium reported in mg per 100 g. • There are a few questions that you might ask:• If there are 1.50 × 103 mg of NaCl per 100 g of chips, how mg of Na are there

per 100 g of chips?• How would the presence of other chloride salts (e.g. KCl) impact calculations of

sodium content based on a silver chloride precipitation reaction?• Other food additives also contain sodium (e.g. monosodium glutamate or

sodium bicarbonate). Would a AgCl precipitation identify other sources of sodium?

Question 2: Determining sulfate and sulfur content in fertiliser

A student took a 0.998 g sample of fertiliser and added it to 100 mL of water. The solution was mixed well until all soluble fertiliser had dissolved. The solution was then filtered. The filtrate was acidified with dilute hydrochloric acid, then treated with excess barium chloride. The precipitate was filtered, dried to a constant mass and weighed. The weight of the precipitate was 0.387 g. Calculate the percentage of sulfate and the percentage of sulfur in the sample of fertiliser.

Where do we start?• Write out the relevant chemical reaction:

SO42-(aq) + Ba2+(aq) BaSO4(s)

• Identify what is important and what is not for the calculation:Is the volume of water added important? (more about this later)Is the presence of HCl important? (more about this later)

Question 2: Determining sulfate and sulfur content in fertiliser

A student took a 0.998 g sample of fertiliser and added it to 100 mL of water. The solution was mixed well until all soluble fertiliser had dissolved. The solution was then filtered. The filtrate was acidified with dilute hydrochloric acid, then treated with excess barium chloride. The precipitate was filtered, dried to a constant mass and weighed. The weight of the precipitate was 0.3827 g. Calculate the percentage by mass of sulfate and the percentage by mass of sulfur in the sample of fertiliser.

Where do we start (continued)?• Identify what we know:

The final mass of BaSO4 is 0.3827 g (4 significant figures) taken from 0.998 g (3 significant figures) of fertiliser

• Think about where you want to get to:Calculate how many grams of sulfate are present in 100 grams of fertiliser. Do the same for sulfur.

Calculating the moles of BaSO4

• SO42-(aq) + Ba2+(aq) BaSO4(s)

m(BaSO4) = 0.3827 gM(BaSO4) = (137.3 + 32.1 + 4 × 16.0) gmol-1

= 233.4 gmol-1

n(BaSO4)= 0.3827 g/233.4 gmol-1

= 0.001640 mol

Calculating the mass of SO42-

• SO42-(aq) + Ba2+(aq) BaSO4(s)

n(BaSO4)= 0. 001640 molFrom the chemical equation we know that the moles of SO4

2- to the moles of BaSO4 is 1:1, therefore:n(SO4

2-) = 0. 001640 molM(SO4

2-) = (32.1 + 4 × 16.0) gmol-1 = 96.1 gmol-1

m(SO42-) = 0. 001640 mol × 96.1 gmol-1 = 0.1576 g

Calculating the percentage by mass of SO4

2- in the fertiliser

(round to 3 sig figs)

The fertiliser contains 15.8% sulfate by mass.

Calculating the percentage by mass of sulfur in the fertiliserThe mole ratio of S to SO4

2- is 1:1.

n(SO42-) = n(S) = 0.001640 mol

m(S) = 32.1 gmol-1 × 0.001640 mol

= 0.05263 g

Are there other considerations?• Is knowing the exact volume of water used to dissolve the sulfate

containing compounds important?• To answer this, consider the purpose of the water and whether it ever came

into play in any of the calculations.

• What is the purpose of the hydrochloric acid?• To answer this, you need a bit more information. • Fertiliser also contains carbonate containing compounds.

• Will carbonate precipitate in the presence of barium ions?• What happens to carbonate in the presence of acid?

Question 3: Determining the empirical formula of a hydrated molecule

Magnesium sulfate readily takes up water and is usually found in a hydrated form with a set number of water molecules per molecule of magnesium sulfate. A student conducted an experiment to calculate the empirical formula for hydrated magnesium sulfate.• Mass of empty crucible: 45.39 g• Mass of crucible, cover and magnesium sulfate before heating: 50.14 g• Mass of crucible, cover and magnesium sulfate after 1st heating: 47.70 g• Mass of crucible, cover and magnesium sulfate after 2nd heating: 47.63 g• Mass of crucible, cover and magnesium sulfate after 3rd heating: 47.62 g

Where do we start?• Think about where you want to get to mole ratio of water to magnesium sulfate

Calculate the number of moles of water• m(H2O) = starting mass – dried mass

= 50.14 g – 47.62 g= 2.52 g

• M(H2O) = (2 × 1.0 + 16.0) gmol-1

= 18 gmol-1

• n(H2O) = 2.52 g/18 gmol-1

= 0.1400 mol

Calculate the number of moles of anhydrous magnesium sulfate• m(MgSO4) = dried mass – crucible

= 47.62 g – 45.18 g= 2.44 g

• M(MgSO4) = (24.3 + 32.1 + 4 × 16.0) gmol-1

= 120.4 gmol-1

• n(MgSO4) = 2.44 g/120.4 gmol-1

= 0.0203 mol

Calculate the mole ratio and determine the empirical formula

The empirical formula hydrated magnesium sulfate is MgSO47H2O

Question 4: Determining the empirical formula of an alcohol

When 12 g of an unknown alcohol was fully combusted, it was found to produce 26.4 g of carbon dioxide and 14.4 g of water. What is the empirical formula for the alcohol?Where do we start?• Write out the relevant chemical reaction:

CxHyOz+ O2 CO2 + H2OWe know that alcohols contain carbon, hydrogen and oxygen, but we don’t know the numbers of each, so we will just call them x, y and z.

• Think about what you know and how you can use it to get started:Knowing the mass of the products will allow the calculation of the number of moles of the products, which will give the number of moles of carbon and hydrogen. Oxygen will be a bit more tricky.

Determining the number of moles of carbon and hydrogen

CxHyOz + O2 CO2 + H2O

Mass (g): 26.414.4Molar mass (gmol-1):44.0 18.0Moles (g/gmol-1 ): 0.600.80Moles (C or H) 0.601.60(note: the number of moles of carbon equals the number of moles of CO2; the number of moles of hydrogen is twice the number of moles of H2O)

Determining the number of moles of oxygen

CxHyOz + O2 CO2 + H2O

Knowing the number of moles of carbon and hydrogen, we can work out the mass of carbon and hydrogen in the alcohol and from that derive the mass of oxygen in the alcohol.

Determining the number of moles of oxygen

CxHyOz + O2 CO2 + H2O

Moles of C and H(mol) 0.6 1.6Molar mass (gmol-1) 12.01.0Mass (mol × gmol-1) 7.2 1.6Mass of oxygen = 12 g – 7.2 g – 1.6 g = 3.2 gMoles of oxygen in the alcohol = 3.2 g/16.0 gmol-1

= 0.2 mol

Determining the empirical formula

CxHyOzMole ratio: 0.6:1.6:0.2Divide through by lowest number in ratio:

3:8:1The empirical formula for the alcohol is C3H8O.

C3H8O – propyl alcohol (IUPAC name: propan-1-ol)