Percentage % element by mass to calculate the composition of a compound and other % composition calculations

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

Keywords: Quantitative Chemistry calculations online Help for problem solving in doing percent of mass of elements in a given compound formula calculations. Practice revision questions on % composition of an element in compound, using experiment data, making predictions. This page describes, and explains, with fully worked out examples, how to calculate the composition of a compound in terms of the % by mass of each element in a compound. It doesn't matter what the nature of the compound is i.e. it is irrelevant whether its an ionic compound or a covalent compound. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses. These revision notes and practice questions on how to do percentage by mass of elements in a compound calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

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type in answer for F and H tiers  or  multiple choice for F and H tiers

The 'percent' % by mass composition of a compound in terms of its constituent elements is calculated in three easy steps

Chemistry calculations 4.

How do you calculate the percent (%) by mass of an element in a compound formula?

How do you calculate the percent (%) by mass of water or an ion in a compound formula?

see section 2. 2. Calculating relative formula/molecular mass (M_r) of a compound

% by mass of Z = 100 x A_r(Z) x atoms of Z / M_r(compound)

It always seems complicated when stated in this formal way, but the calculations are actually quite easy ..

as long as you can correctly read a formula!

• Calculation of % composition Example 4a.1
  • Calculate the % of copper in copper sulphate, CuSO₄
  • Relative atomic masses: Cu = 64, S = 32 and O = 16
  • relative formula mass = 64 + 32 + (4x16) = 160
  • only one copper atom of relative atomic mass 64
  • % Cu = 100 x 64 / 160
  • = 40% copper by mass in the compound
    • Note that similarly, you can calculate the % of the other elements in the compound e.g.
    • % sulfur = (32/160) x 100 = 20% S
    • % oxygen = (64/160) x 100 = 40% O
    • Also note that if you haven't made any errors, they should add up to 100%,useful arithmetical !
• Calculation of % composition Example 4a.2
  • Calculate the % of oxygen in aluminium sulphate, Al₂(SO₄)₃
  • Relative atomic masses: Al = 27, S = 32 and O = 16
  • relative formula mass = 2x27 + 3x(32 + 4x16) = 342
  • there are 4 x 3 = 12 oxygen atoms, each of relative atomic mass 16,
  • giving a total mass of oxygen in the formula of 12 x 16 = 192
  • % O = 100 x 192 / 342 = 56.1% oxygen by mass in aluminium sulphate
• Calculation of % composition Example 4a.3
  • The next two examples extend the idea of % element composition to include % composition of part of a compound, in these cases water in a hydrated salt and the sulfate ion in a potassium salt.
  • Calculate the % of water in hydrated magnesium sulphate MgSO₄.7H₂O
  • Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1
  • relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246
  • 7 x 18 = 126 is the mass of  water
  • so % water = 100 x 126 / 246 = 51.2 % H₂O
  • Note: The determination and calculation of the formula of a hydrated salt like MgSO₄.7H₂O is covered in Calculations section 14.4.
• Calculation of % composition Example 4a.4
  • Calculate the percentage by mass, of sulfate ion in sodium sulfate
  • formula of sodium sulfate Na₂SO₄, atomic masses: Na = 23, S = 32, O = 16
  • Formula mass Na₂SO₄ = (2 x 23) + 32 + (4 x 16) = 142
  • Formula mass of sulfate ion SO₄^2- (or just SO₄ will do for the calculation) = 32 + (4 x 16) = 96
  • Therefore % sulfate ion in sodium sulfate = (96/142) x 100 = 67.6% SO₄

Atomic masses used for 4b. questions: C = 12, Cl = 35.5, Fe = 56, H = 1, Mg = 24, N = 14, Na = 23, O = 16, S = 32,

By now I assume you can do formula mass calculations and read formula without any trouble, so ALL the detail of such calculations is NOT shown, just the bare essentials!

Example 4b.1

Ammonium sulfate, (NH₄)₂SO₄, is an important ingredient in many artificial fertilisers supplying to plants the essential mineral elements of nitrogen and sulfur.

(a) Calculate the percentage of nitrogen and the percentage of sulfur in ammonium sulfate.

formula mass of ammonium sulfate = (2 x 18) + 32 + 64 = 132.

with two nitrogen atoms in the formula; % nitrogen by mass  = 100 x 28/132 = 21.2% N

with one sulfur atom in the formula, % sulfur by mass = 100 x 32/132 = 24.2% S

(b) Calculate the percentage of sulfate ion in ammonium sulfate.

To calculate the percentage of a 'part' of a compound, you just use the formula mass of that 'part!

formula mass of sulfate, SO₄, is 32 + (4 x 16) = 96

therefore % sulfate by mass = 100 x 96/132 = 72.7% SO₄

Note: If the question refers to the sulfate ion itself, SO₄^2-, its just the same % mass calculation!

Example 4b.2

What is the percentage of carbonate ion in sodium carbonate? (Na₂CO₃)

formula mass of sodium carbonate = 46 + 12 + 48 = 106

formula mass of carbonate, CO₃ = 12 + (3 x 16) = 60

therefore % carbonate ion by mass = 100 x 60/106 = 56.6% CO₃ (for the CO₃^2- ion)

Example 4b.3

Calculate the percentage water of crystallisation in magnesium sulfate crystals, MgSO₄.7H₂O, known as Epsom salt.

formula mass of Epsom salt = 24 + 32 + 64 + (7 x 18) = 246

formula mass of water = 18, mass of seven water molecules is 7 x 18 = 126

therefore % of water of crystallisation in the crystals = 100 x 126/246 = 51.2% H₂O

Example 4b.4

Rock salt is mainly sodium chloride, NaCl

On analysis of an impure sample of rock salt, it was found to contain by mass 57.5% of chlorine as chloride ion.

(a) Calculate the percentage purity of the salt.

the formula mass of sodium chloride is 58.5

the formula mass of chloride is 35.5

therefore you need to scale up from the % mass of chloride ion to the % mass of sodium chloride.

the scale up factor must be 58.5/35.5 = 1.648

therefore percentage of sodium chloride in the rock salt = 57.5 x 1.648 = 94.8% NaCl

(b) What assumption have you made in this calculation to make this a valid calculation?

You have assumed that non of the impurities contain the sodium or chloride ion.

There may be other sodium or chloride salts in the rock salt mixture.

Example 4b.5

A mixture of sand and a compound based on iron(II) sulfate (*), FeSO₄. is used to treat grass e.g. lawns and bowling greens to promote plant growth and kill moss.

What percentage by mass of iron(II) sulfate is required in the mixture to give 15% by mass of iron(II) ions (Fe²⁺)?

You need to scale up from the mass of iron ions to the mass of the compound FeSO₄.

formula mass of FeSO₄ = 56 + 32 + 64 = 152

atomic mass of iron Fe or iron(II) ion Fe²⁺ = 56 (note the atom and ion have the same mass!)

therefore the scaling up factor is 152/56 = 2.714

therefore % iron(II) sulfate required in the mixture = 15 x 2.714 = 40.7% FeSO₄

Note (*): The actual iron compound used in lawn treatments is crystals of ammonium iron(II) sulfate,

(NH₄)₂Fe(SO₄)₂.6H₂O, old name ferrous ammonium sulphate, its a double salt, but I've just based the calculation on the iron(II) sulfate part.

Example 4b.6

A baking powder mixture contains sodium hydrogencarbonate, NaHCO₃.

To get sufficient rising action from the carbon dioxide gas (CO₂) formed in baking, it should contain a minimum of 50% carbonate ion (CO₃^2-).

Calculate the minimum percentage of sodium hydrogen carbonate that should be in the mixture.

You need to scale up from the formula masses of the carbonate ion and that of sodium hydrogen carbonate.

formula mass of carbonate, CO₃ = 12 + 48 = 60 (same for the carbonate ion)

formula mass sodium hydrogencarbonate = 23 + 1 + 60 = 84.

therefore the scale up factor is 84/60 = 1.4

so, minimum percentage sodium hydrogen carbonate in the mixture should be 50 x 1.4 = 70% NaHCO₃

 

Example 4b.7

In an experiment 6.0 g of metal M was burned in a crucible, by heating in air, until there was no more gain in weight. Apart from the mass of the crucible, the final mass of the residue was 10.0 g. The oxide O formed was an essential ingredient in a ceramic pigment mixture P for glazing pottery.

(a) What % of the oxide O is the metal M.

100 x 6.0/10.0 = 60% of M in the oxide O

(b) The mixture P must contain 25% by mass of the metal M.

What mass of the oxide O is needed to make 12 g of the mixture P?

25% of 12 g is 12 x 25/100 = 3.0 g, so this is the mass of metal M in 12 g of mixture P as the oxide O

You now have to scale up to the mass of the oxide O needed which contains 60% of M.

Scaling up gives 3.0 x 100/60 = 5.0 g of oxide M is needed.

PLEASE NOTE:

Above is typical periodic table used in science-chemistry courses for use in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

However, for calculations of percentage composition (and any other quantitative chemistry calculations) note:

(i) At GCSE level, relative atomic masses are quoted as whole numbers (integers) e.g. C = 12, Fe = 56, Ag = 108 etc.

apart from copper Cu = 63.5 and chlorine Cl = 35.5

(ii) In advanced A level chemistry, (but still pre-university) relative atomic masses will be quoted to one decimal place

e.g. H = 1.0, C = 12.0, Cl = 35.5, Fe = 55.8, Cu = 63.5, Ag 107.9 etc.

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OTHER CALCULATION PAGES

1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

2. Calculating relative formula/molecular mass of a compound or element molecule

3. Law of Conservation of Mass and simple reacting mass calculations

4. Composition by percentage mass of elements in a compound (this page)

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

   • Reacting masses, concentration of solution and volumetric titration calculations (NOT using moles)

7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

13. Electrolysis products calculations (negative cathode and positive anode products)

14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

    • 14.1 % purity of a product 14.2a % reaction yield 14.2b atom economy 14.3 dilution of solutions

    • 14.4 water of crystallisation calculation  14.5 how much of a reactant is needed? limiting reactant

15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

17. Radioactivity & half-life calculations including dating materials

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