Find the ratio by mass of the combining elements in the following compounds. (You may use Appendix $-III$)
$(a)$ $CaCO_{3}$ $(b)$ $MgCl_{2}$
$(c)$ $H_{2}SO_{4}$ $(d)$ $C_{2}H_{5}OH$
$(e)$ $NH_{3}$ $(f)$ $Ca(OH)_{2}$
$(a)$ $CaCO_{3}$ $(b)$ $MgCl_{2}$
$(c)$ $H_{2}SO_{4}$ $(d)$ $C_{2}H_{5}OH$
$(e)$ $NH_{3}$ $(f)$ $Ca(OH)_{2}$
(N/A) $Ca : C : O_{3} = 40 : 12 : (16 \times 3) = 40 : 12 : 48 = 10 : 3 : 12$
$(b)$ $Mg : Cl_{2} = 24 : (35.5 \times 2) = 24 : 71$
$(c)$ $H_{2} : S : O_{4} = (1 \times 2) : 32 : (16 \times 4) = 2 : 32 : 64 = 1 : 16 : 32$
$(d)$ $C_{2} : H_{6} : O = (12 \times 2) : (1 \times 6) : 16 = 24 : 6 : 16 = 12 : 3 : 8$
$(e)$ $N : H_{3} = 14 : (1 \times 3) = 14 : 3$
$(f)$ $Ca : O_{2} : H_{2} = 40 : (16 \times 2) : (1 \times 2) = 40 : 32 : 2 = 20 : 16 : 1$
$(b)$ $Mg : Cl_{2} = 24 : (35.5 \times 2) = 24 : 71$
$(c)$ $H_{2} : S : O_{4} = (1 \times 2) : 32 : (16 \times 4) = 2 : 32 : 64 = 1 : 16 : 32$
$(d)$ $C_{2} : H_{6} : O = (12 \times 2) : (1 \times 6) : 16 = 24 : 6 : 16 = 12 : 3 : 8$
$(e)$ $N : H_{3} = 14 : (1 \times 3) = 14 : 3$
$(f)$ $Ca : O_{2} : H_{2} = 40 : (16 \times 2) : (1 \times 2) = 40 : 32 : 2 = 20 : 16 : 1$
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