Test Index

WAEC Chemistry 2022 Paper

Question : 32

Total: 50

\mathrm{Ca}(\mathrm{HCO}_3)_2

[\mathrm{Ca}=40.0,\ \mathrm{O}=16.0,\ \mathrm{C}=12.0,\ \mathrm{H}=1.0]

Solution:

Concept:

Percentage composition is the mass of a particular element in one mole of the compound, divided by the total molar mass, multiplied by 100.

Explanation:

Molar mass of

\mathrm{Ca}(\mathrm{HCO}_3)_2

is:

40 + 2[1 + 12 + 3(16)]

= 40 + 2[13 + 48]

= 40 + 2(61) = 40 + 122 = 162

Mass of carbon in one mole of the compound is:

2 \times 12 = 24

Percentage composition of carbon is:

\frac{24}{162} \times 100 = 14.8\%

Answer:

14.8% (Option B).

Go to Question:

Prev Question

Next Question