Percentage Composition of Elements | Study Planner | Revision Questions | Target Teach Tutors
In this video, we simplify the concept of percentage composition of an element in chemistry with detailed practical examples and revision questions in the description. Covering all the essential points for your GCSE Chemistry exam, this video is a must-watch to help you ace your exams!
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Answers to the questions.
Question 1:
Ans: H -Number of elements 2 and the total mass is 2 x 1 = 2
O - Number of elements 1 and the total mass is 1 x 16 = 16
Total mass of H₂O is 2 + 16 = 18
% of composition of H is = Total mass of H ÷ Total mass of H₂O x 100
= 2 ÷ 18 x 100 = 11.11%
% of composition of O is = 100 – 11.11 = 88.89%
Question 2:
Ans: Na -Number of elements is 2 and the total mass is 2 x 23 = 46
C - Number of elements is 1 and the total mass is 1 x 12 = 12
O - Number of elements is 3 and the total mass is 3 x 16 = 48
Total mass of Na₂CO₃ is 46 + 12 + 48 = 106
% of composition of Na is = Total mass of Na ÷ Total mass of Na₂CO₃ x 100
= 46 ÷ 106 x 100 = 43.39%
% of composition of C is = Total mass of C ÷ Total mass of Na₂CO₃ x 100
= 12 ÷ 106 x 100 = 11.32%
% of composition of O is = 100 – (43.39 + 11.32) = 45.29%
Question 3:
Ans: Mg -Number of elements 1 and the total mass is 1 x 24 = 24
O - Number of elements 1 and the total mass is 1 x 16 = 16
Total mass of MgO is 24 + 16 = 40
% of composition of Mg is = Total mass of Mg ÷ Total mass of MgO x 100
= 24 ÷ 40 x 100 = 60%
% of composition of O is = 100 – 60 = 40%
Question 4:
Ans: Na -Number of elements is 1 and the total mass is 1 x 23 = 23
H – Number of elements is 1 and the total mass is 1 x 1 = 1
C - Number of elements is 1 and the total mass is 1 x 12 = 12
O - Number of elements is 3 and the total mass is 3 x 16 = 48
Total mass of NaHCO₃ is 23 + 1 + 12 + 48 = 84
% of composition of Na is = Total mass of Na ÷ Total mass of NaHCO₃ x 100
= 23 ÷ 84 x 100 = 27.4%
% of composition of H is = Total mass of H ÷ Total mass of NaHCO₃ x 100
= 1 ÷ 84 x 100 = 1.2%
% of composition of C is = Total mass of C ÷ Total mass of NaHCO₃x 100
= 12 ÷ 84 x 100 = 14.3%
% of composition of O is = 100 – (27.38 + 1.19 + 14.29) = 57.1%
Question 5:
Ans: Ca - Number of elements 1 and the total mass is 1 x 40 = 40
C - Number of elements 1 and the total mass is 1 x 12 = 12
O - Number of elements 3 and the total mass is 3 x 16 = 48
Total mass of CaCO₃ is 40 + 12 + 48 = 100
% of composition of Ca is = Total mass of Ca ÷ Total mass of CaCO₃ x 100
= 40 ÷ 100 x 100 = 40%
% of composition of C is = Total mass of C ÷ Total mass of CaCO₃ x 100
= 12 ÷ 100 x 100 = 12%
% of composition of O is = Total mass of O ÷ Total mass of CaCO₃ x 100
= 48 ÷ 100 x 100 = 48%
Question 6:
Ans: C - Number of elements 6 and the total mass is 6 x 12 = 72
H - Number of elements 1 and the total mass is 1 x 12 = 12
O - Number of elements 6 and the total mass is 6 x 16 = 96
Total mass of C₆H₁₂O₆ is 72 + 12 + 96 = 180
% of composition of C is = Total mass of C ÷ Total mass of C₆H₁₂O₆ x 100
= 72 ÷ 180 x 100 = 40%
% of composition of H is = Total mass of H ÷ Total mass of C₆H₁₂O₆ x 100
= 12 ÷ 180 x 100 = 6.7%
% of composition of O is = Total mass of O ÷ Total mass of C₆H₁₂O₆ x 100
= 96 ÷ 180 x 100 = 53.3%
Question 7:
Ans: Ca - Number of elements 3 and the total mass is 3 x 40 = 120
P - Number of elements 2 and the total mass is 2 x 31 = 62
O - Number of elements 8 and the total mass is 8 x 16 = 128
Total mass of Ca₃(PO₄)₂ is 120 + 62 + 128 = 310
% of composition of Ca is = Total mass of Ca ÷ Total mass of Ca₃(PO₄)₂ x 100
= 120 ÷ 310 x 100 = 38.7%
% of composition of P is = Total mass of P ÷ Total mass of Ca₃(PO₄)₂ x 100
= 62 ÷ 310 x 100 = 20%
% of composition of O is = Total mass of O ÷ Total mass of Ca₃(PO₄)₂ x 100
= 128 ÷ 310 x 100 = 41.3%
Study hard!