Empirical formula notes and questions Answers PDF

Preview

Summary

Lecture Notes with Q&A...

Description

EMPIRICAL AND MOLECULAR FORMULAE EMRIRICAL FORMULA – This represents the simplest whole number ratio of atoms within a molecule MOLECULAR FORMULA – This represents the exact numbers of each atom in a molecule Empirical formulae can be calculated from either % mass composition data or mass composition data. Molecular formula can then be calculated if the molecular mass of the compound is known. EXAMPLE 1 – from % composition A substance containing carbon and hydrogen and oxygen only contains 73.47% carbon and 10.20% hydrogen by mass, its relative molecular mass is 98. Calculate its molecular formula. Step 1 calculate empirical formula ATOMS CARBON PERCENTAGE MASS 73.47 ATOMIC MASS MASS/ATOMIC MASS RATIO(divide by smallest)

12 6.1225 6

Empirical formula = C6H10O Step 2 use MR to find molecular formula MR = mass (C6H10O) x n 98 = (98) x n C6H10O

HYDROGEN 10.20 1 10.20 10

n = 1

OXYGEN (100 – (73.47 + 10.20) = 16.33

16 1.020 1

molecular formula =

EXAMPLE 2 – from mass data A substance contains 12.28g nitrogen, 3.51g hydrogen, 28.07g sulphur and 56.10g oxygen, its MR is 228 .Calculate its molecular formula.

Step 1 calculate empirical formula ATOMS N MASS 12.28 ATOMIC MASS 14 %MASS/ATOMIC 0.877 MASS RATIO(divide by 1 smallest)

H 3.51 1 3.51

S 28.07 32 0.877

O 56.10 16 3.506

4

1

4

Empirical formula = NH4SO4 Step 2 use MR to find molecular formula MR = mass (NH4SO4) x n 228 = (114) x n n=2

moloecular formula = N2H8S2O8

QUESTION 1 A substance contains 40% carbon, 6.67%hydrogen and 53.3% oxygen, its MR is 60. Find its molecular formula. Step 1 calculate empirical formula ATOMS C H O PERCENTAGE MASS 40 6.67 53.3 ATOMIC MASS 12 1 16 %MASS/ATOMIC 40/12 6.67/1 53.3/16 MASS =3.33 =6.67 =3.33 RATIO(divide by 1 2 1 smallest) Empirical formula = CH2O Step 2 use MR to find molecular formula MR = mass ( 30 ) x n 60= ( 30 ) x n, n=2

molecular formula = C2H4O2

QUESTION 2 A substance contains 43.66g phosphorous and 56.34g oxygen, its MR is 284. Find its molecular formula.

Step 1 calculate empirical formula ATOMS P MASS 43.66g ATOMIC MASS 31 MASS/ATOMIC MASS 43.66/31=1.41 RATIO(divide by 1 smallest) Empirical formula = P2O5 Step 2 use MR to find molecular formula MR = mass ( 142 ) x n 284= (142) x n, n=2

O 56.34g 16 56.34/16=3.52 2.5

molecular formula = P4O10

QUESTION 3 - Calculate the empirical and molecular formulae for the following compounds showing all of your workings. (a) Lead 92.8% and Oxygen 7.2%, Mr = 446 (b) Hydrogen 5g, nitrogen 35g and oxygen 60g, Mr = 80 (c) Phosphorous 10.88% and Iodine 89.12%, Mr = 570 You may be provided with experimental data and asked to calculate empirical and molecular formulae EXAMPLE 3 Analysis of a hydrocarbon showed that 7.8g of the hydrocarbon contained 0.6g of hydrogen and that the Mr = 78. What is the formula of the hydrocarbon? ATOMS MASS ATOMIC MASS MASS/ATOMIC MASS RATIO(divide by smallest)

C H 7.8 – 0.6 = 7.2 0.6 12 1 0.6 0.6 1 1

Empirical formula = CH Step 2 use MR to find molecular formula MR = mass ( CH ) x n

78 = (13 ) x n,

n=6

molecular formula = C6H6

EXAMPLE 4 22.3g of an oxide of lead produced 20.7 grams of metallic lead on reduction with hydrogen. Calculate its empirical formula. ATOMS MASS

Pb 20.7

ATOMIC MASS MASS/ATOMIC MASS RATIO(divide by smallest)

207 0.1 1

O 22.3 – 20.7 = 1.6 16 0.1 1

Empirical formula = PbO QUESTION 4 3.36g of iron join with 1.44g of oxygen in an oxide of iron that has a Mr = 160. What is the molecular formula of the oxide? ATOMS MASS ATOMIC MASS

Fe 3.36g 56

O 1.44g 16

MASS/ATOMIC MASS RATIO(divide by smallest)

0.06 1

0.09 1.5

Empirical formula = Fe2O3 Step 2 use MR to find molecular formula MR = mass ( 160 )xn 160 = ( 160 ) x n, n=

1

molecular formula =Fe2O3

Question 5 When 1.17g of potassium is heated in oxygen, 2.13g of potassium oxide is produced. What is the empirical formula for this oxide? ATOMS MASS ATOMIC MASS %MASS/ATOMIC MASS

K 1.17g 39 0.03

O 0.96g 16 0.06

RATIO(divide by smallest)

1

2

Empirical formula = KO2 QUESTION 6 (a) a hydrocarbon contains 92.3% carbon and has a relative mass of 26. Find its molecular formula. (C2H2) (b) Sodium burns in excess oxygen to give a yellow solid oxide that contains 58.97% of sodium. What is the empirical formula of the oxide? (NaO) ANHYDROUS AND HYDRATED SOLIDS ANHYDROUS – this means without water – contains no waters of crystalisation HYDRATED – this means contains water – ie for ionic solids contains waters of crystalisation EXAMPLE 5 When 14.97g of hydrated copper II sulphate is heated it produces 9.60g of anhydrous copper II sulphate. What is the formula of the hydrated salt.   

In order to complete this calculation we must work out the formula for copper II sulphate – CuSO4. We must then calculate the mass of water in the hydrated sample ( 14.97 – 9.6 = 5.37g). We can then use this mass, along with the mass of anhydrous copper II sulphate to work out the proportion of moles of each in the crystal.

COMPOUNDS MASS MOLECULAR MASS MASS/ATOMIC MASS RATIO(divide by smallest)

CuSO4 9.6 159.5 0.06 1

H2O 5.37 18 0.30 5

Formula = CuSO4 . 5H2O QUESTION 7 24.6g of a hydrated salt of MgSO4 . x H2O, gives 12.0g of anhydrous MgSO4 on heating. What is the value of x? Calculate the mass of water evolved =

hydrated mass – anhydrous mass = 12.6g

Mass of water evolved = 12.6 g COMPOUNDS MASS MOLECULAR MASS

MgSO4 12.0g 120

H2O 12.6g 18

MASS/ATOMIC MASS RATIO(divide by smallest)

0.1 1

0.7 7

X = 7 QUESTION 8 (a) 3.715g of zinc sulphate crystals were heated until no further heat loss. The anhydrous solid had a mass of 2.086g. Find x in ZnSO4 . x H2O x=6 (b) When 14.2g of hydrated nickel II nitrate is dried, 8.9g remain. Find x in Ni(NO3)2 . x H2O x=6...