2.1.3 b,c,d Determination of Formula

Syllabus

Determination of formulae

(b) use of the terms:

(i) empirical formula (the simplest whole number ratio of atoms of each element present in a compound) - Definition not required.

(ii) molecular formula (the number and type of atoms of each element in a molecule) - Definition not required.

(c) calculations of empirical and molecular formulae from composition by mass or percentage compositions by mass and relative molecular mass - To include calculating empirical formulae from elemental analysis data

(d) the terms anhydrous, hydrated and water of crystallisation and calculation of the formula of a hydrated salt from given percentage composition, mass composition or based on experimental results

Empirical and Molecular Formulae

These have been discussed in a previous section.

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GCSE Revision

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If you have any doubt how to work out ionic formula view this first

Empirical Formula Calculations

i) Empirical and molecular formula from mass

Simple

You should already be familiar with the idea that we can calculate the number of moles of a pure substance using the formula:

Moles = Mass ÷ Molar Mass

So if we know the mass of elements within a substance

we can work out the number of moles of each element
then we can work out their simplest integer ratio
then we can write an empirical formula
and if we also know the M_r we can work out the molecular formula

For instance,

A substance has M_r = 58 g/mol. It is found to contain 9.6g of Carbon and 2.0g of Hydrogen only.

Find its empirical formula and hence its molecular formula.

Element Carbon Hydrogen

Mass(g) 9.6 2.0

Moles 9.6 ÷ 12 = 0.8 2.0 ÷ 1.0 = 2.0

Divide by smallest 0.8 ÷ 0.8 = 1 2.0 ÷ 0.8 = 2.5

Not integer so double all 2 5

Empirical formula = C₂H₅

M_r (C₂H₅) = (12 x 2) + (1 x 5) = 29

From the question the actual M_r = 58

58 ÷ 29 = 2 So there are two lots of the empirical formula in the molecular formula

Molecular formula = C₄H₁₀

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More Complicated

Copper Oxide can be reduced by passing gas over it.

You could be given results like the ones below.

Mass of empty tube (g)

2.25

Mass of tube & Copper Oxide (g)

6.42

Mass of tube & Copper (g)

6.06

1.Use the results to work out the masses of each element

Mass of Copper Oxide = 6.42 - 2.25 = 4.17g

Mass of Copper = 6.06 - 2.25 = 3.81g

Mass of Oxygen removed = 4.17 - 3.81 = 0.36g

2. Use the same technique as above

Element Copper Oxygen

Mass(g) 3.81 0.36

Moles 3.81 ÷ 63.5 = 0.06 0.36 ÷ 12 = 0.03

Divide by smallest 0.06 ÷ 0.03 = 2 0.03 ÷ 0.03 = 1

Empirical formula = Cu₂O

ii) Empirical Formula from Percentage Composition

For example a substance is investigated and found to have an M_r= 46 g/mol.

It contains 52.17% Carbon, 13.04% Hydrogen and the rest in Oxygen.

Find its empirical formula and hence its molecular formula.

The key is to treat the percentages as masses. Then use the same technique.

Element C H O

Mass(g) 52.17 13.04 34.78

Moles 52.17 ÷ 12 = 4.35 13.04 ÷ 1 = 13.04 34.78 ÷ 16 = 2.17

Divide by smallest 4.35 ÷ 2.17 = 2 13.04 ÷ 2.17 = 6 2.17 ÷ 2.17 = 1

Empirical Formula = C₂H₆O

M_r (C₂H₆O) = 46

This is the same as the molecular M_r so the empirical and molecular formulas are the same.

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Water of Crystallisation Calculations

If we heat a hydrated salt until it is anhydrous (without water) then we can use the masses to work out the ratio of water to salt.

eg.

Mass of empty crucible (g)

19.0

Mass of Crucible & hydrated salt (g)

25.25

Mass of Crucible & anhydrous salt (g)

23.0

Mass of hydrated Copper(II) Sulphate, CuSO₄.xH₂O = 25.25 - 19.0 = 6.25g
Mass of anhydrous salt = 23.0 - 19.0 = 4.00g
Mass of water removed = 6.25 - 4.00 = 2.25g
Moles of Anhydrous salt = Mass / Mr (CuSO₄) = 4 / 160 = 0.025
Moles of water removed = Mass / Mr (H₂O) = 2.25 / 18 = 0.125
The mass ratio of CuSO₄ : H₂O = 0.025 : 0.125 = 1:5
Empirical formula = CuSO₄.5 H₂O

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Exam-style Questions

1. (a) Give the meaning of the term empirical formula

…………………………………………………………………...............................……...................................................................……………………......………………………………………..........................……………………………………………. …….. (1)

(b) Analysis of 3.150 g of compound X showed that it contained 0.769 g of calcium and 0.539 g of nitrogen; the remainder was oxygen. Calculate the empirical formula of X.

………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

(3)

(c) What additional information is required in order to deduce the molecular formula of X.

…………………………………………………...............................................................……………………………….. (1)

2.

(a) Give the meaning of the term empirical formula.

...............................................................................................................................................................................................

...............................................................................................................................................................................................(1)

(b) Define the term relative molecular mass.

...............................................................................................................................................................................................

...............................................................................................................................................................................................(2)

(c) The empirical formula of a compound is CHO and its relative molecular mass has the value 174. Determine the molecular formula of this compound and show your working.

...............................................................................................................................................................................................

.............................................................................................................................................................................................(2)

(d) A compound with molecular formula CH4O burns in air to form carbon dioxide and water. Write a balanced equation for this reaction.

..............................................................................................................................................................................................(1)

(Total 6 marks)

Answers

1

(a) simplest ratio of atoms of each element in a compound (1) (1)

(b) mass of O = 1.842 g (1)

Ca : N : O = 0.769/40 : 0.539/14 : 1.842/16 (1)

= 1:2:6

CaN₂O₆ (1)

(c) Mr (1)

2.

(a) simplest ratio of atoms of each element in a compound (1)

(b) (Average mass of substance)/(Mass of one atom of ¹²C) (1)

× 12 (1)

(or definition in terms of moles)

(c) CHO = 29 174/29= 6 (1)

\ C6 H6 O6 (1)

(d) CH₄O+ 1.5 O₂ ¾® CO2+2H2O (1)