Moles, Molar Mass and Empirical Formula Calculations - Worksheets, Questions and Revision

9 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

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This topic is chapter 9 of IGCSE Chemistry Practice Book.

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2.9 Moles, Molar Mass and Empirical Formula Calculations

EDEXCEL 4CH1 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer in full sentences for Questions 1 and 2 only. Use a calculator where allowed. Suggested time 60 minutes.

State the value of the Avogadro constant used in the Edexcel IGCSE context, and give its units.

(Total for Question 1 is 1 mark)

A sample contains 0.250 mol of sodium chloride, NaCl. Calculate the mass in grams of this sample. Use Ar: Na = 23.0, Cl = 35.5. Quote the equation, substitute and give the final answer with units.

(Total for Question 2 is 4 marks)

Calculate the number of moles in 24 g of carbon dioxide, CO2. Use Ar: C = 12, O = 16 and show the equation, substitution and final evaluation with units.

(Total for Question 3 is 3 marks)

A 10.0 g sample of magnesium oxide was analysed and found to contain 6.12 g of magnesium and 3.88 g of oxygen. Determine the empirical formula of this magnesium oxide. Show the steps: convert masses to moles, find simplest whole number ratio and give the formula.

(Total for Question 4 is 3 marks)

An organic compound is analysed and found to contain 40.0% carbon, 6.71% hydrogen and 53.29% oxygen by mass. Determine the empirical formula of the compound. Show working: assume 100 g sample, convert to moles, find simplest whole number ratio and give empirical formula.

(Total for Question 5 is 4 marks)

A volatile compound has empirical formula CH2 and an experimentally determined relative molecular mass Mr of 56.0. Calculate the molecular formula. Show the equation, substitution and final result.

(Total for Question 6 is 3 marks)

A student combusts 0.500 g of an unknown metal and obtains 0.644 g of the metal oxide product. The mass of oxygen combined is therefore 0.144 g. Using Ar: O = 16.0, and a metal Ar of 27.0, determine the empirical formula of the oxide. Show all steps converting masses to moles and finding simplest ratio.

(Total for Question 7 is 6 marks)

A molecule has empirical formula CH and empirical Mr 78. Calculate the molecular formula. Show the empirical Mr calculation, division step and final molecular formula.

(Total for Question 8 is 5 marks)

A hydrated salt is analysed and found to have formula CuSO4.xH2O. A 2.50 g sample of the hydrated salt was heated to remove water and produced 1.60 g of anhydrous CuSO4. Determine the value of x. Use Ar: Cu = 63.5, S = 32.1, O = 16.0, H = 1.01. Show all steps converting masses to moles and finding whole number x.

(Total for Question 9 is 5 marks)

Mark scheme · 2.9 Moles, Molar Mass and Empirical Formula Calculations

Question 1

B1 6.02 x 10²³ particles per mole (or per mol)
Answer: 6.02 x 10²³ particles per mole

Question 2

M1 writes Mr NaCl = 23.0 + 35.5 = 58.5 and equation mass = moles x Mr or moles = mass / Mr rearranged to mass = moles x Mr
M1 substitutes: mass = 0.250 x 58.5
A1 evaluates mass = 14.625 g
A1 gives final answer with correct units, mass = 14.6 g (to 3 s.f.) cao
Answer: Mr NaCl = 58.5; mass = 0.250 x 58.5 = 14.6 g

Question 3

M1 writes equation Mr CO2 = 12 + (2 x 16) = 44 and moles = mass / Mr
M1 substitutes: moles = 24 / 44
A1 gives moles = 0.545 mol cao (to 3 s.f.)
Answer: Mr CO2 = 12 + 2(16) = 44; moles = 24 / 44 = 0.545 mol

Question 4

M1 converts masses to moles: moles Mg = 6.12 / 24.3 and moles O = 3.88 / 16
M1 divides by smallest to get ratio and simplifies to whole numbers
A1 gives empirical formula MgO cao
Answer: Moles Mg = 6.12 / 24.3 = 0.252 mol; moles O = 3.88 / 16 = 0.243 mol; ratio approx 0.252:0.243 -> 1.04:1 -> empirical formula MgO

Question 5

M1 assumes 100 g: masses C 40.0 g, H 6.71 g, O 53.29 g and converts to moles using Ar 12.0, 1.01, 16.0
M1 calculates moles: C = 40.0/12.0, H = 6.71/1.01, O = 53.29/16.0
M1 divides by smallest mole value and finds simplest whole number ratio
A1 gives empirical formula CH2O cao
Answer: Moles: C = 40.0/12.0 = 3.333; H = 6.71/1.01 = 6.644; O = 53.29/16.0 = 3.330. Ratio C:H:O = 3.333:6.644:3.330 -> divide by 3.330 -> 1.001:1.996:1 -> approx 1:2:1 so empirical formula CH2O

Question 6

M1 calculates Mr of empirical formula CH2 = 12.0 + 2(1.01) = 14.02
M1 divides Mr molecular by empirical Mr: 56.0 / 14.02 and finds integer n
A1 gives molecular formula C4H8 cao
Answer: Mr(empirical) = 12.0 + 2(1.01) = 14.02; n = 56.0 / 14.02 = 4.00; molecular formula = C4H8

Question 7

M1 finds mass of metal and oxygen combined into moles: moles metal = 0.500 / 27.0, moles O = 0.144 / 16.0
M1 calculates numerical mole values correctly
M1 divides both mole values by the smallest to get simplest ratio
M1 adjusts ratio to whole numbers if necessary and states the simplest whole number ratio
A1 gives empirical formula M2O, consistent with the calculated 2:1 metal to oxygen ratio, cao
A1 final empirical formula stated correctly with correct subscripts
Answer: Moles metal = 0.500 / 27.0 = 0.01852 mol; moles O = 0.144 / 16.0 = 0.00900 mol; ratio metal:O = 0.01852:0.00900 = 2.058:1 -> approx 2:1 so empirical formula M2O

Question 8

M1 calculates Mr(empirical) = 12.0 + 1.01 = 13.01
M1 divides 78 by 13.01 to find n
M1 finds integer n = 6 (or 5.993 awrt) and states this
A1 constructs molecular formula by multiplying empirical subscripts by n
A1 gives final molecular formula C6H6 cao
Answer: Mr(empirical) = 13.01; n = 78 / 13.01 = 6.00; molecular formula = C6H6

Question 9

M1 finds mass of water lost = 2.50 - 1.60 = 0.90 g and converts to moles of water: 0.90 / 18.02
M1 calculates Mr CuSO4 = 63.5 + 32.1 + 4(16.0) = 159.6 and converts mass anhydrous to moles: 1.60 / 159.6
M1 computes numerical mole values correctly
M1 divides moles H2O by moles CuSO4 to find x and rounds to nearest whole number
A1 gives integer value x = 5 cao
Answer: Mass H2O = 0.90 g -> moles H2O = 0.90 / 18.02 = 0.04995 mol; Mr CuSO4 = 159.6, moles CuSO4 = 1.60 / 159.6 = 0.01003 mol; x = 0.04995 / 0.01003 = 4.98 approx 5 so x = 5

Moles, Molar Mass and Empirical Formula Calculations - Worksheets, Questions and Revision

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