Difficult
Calculate the percentage composition of the following elements:
$(a)$ Sulphur in $CN_4H_4S$
$(b)$ Carbon in $C_2H_4O_2$
$(c)$ Nitrogen in $CN_2H_2O$
$(d)$ Hydrogen in $C_6H_6$
Given atomic masses: $C = 12, H = 1, O = 16, N = 14, S = 32$.
$(a)$ Sulphur in $CN_4H_4S$
$(b)$ Carbon in $C_2H_4O_2$
$(c)$ Nitrogen in $CN_2H_2O$
$(d)$ Hydrogen in $C_6H_6$
Given atomic masses: $C = 12, H = 1, O = 16, N = 14, S = 32$.
(N/A) Molar mass of $CN_4H_4S = 12 + (4 \times 14) + (4 \times 1) + 32 = 12 + 56 + 4 + 32 = 104 \ g/mol$. Percentage of $S = (32 / 104) \times 100 = 30.77 \%$.
$(b)$ Molar mass of $C_2H_4O_2 = (2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \ g/mol$. Percentage of $C = (24 / 60) \times 100 = 40.0 \%$.
$(c)$ Molar mass of $CN_2H_2O = 12 + 14 + (2 \times 1) + 16 = 12 + 14 + 2 + 16 = 44 \ g/mol$. Percentage of $N = (14 / 44) \times 100 = 31.82 \%$.
$(d)$ Molar mass of $C_6H_6 = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \ g/mol$. Percentage of $H = (6 / 78) \times 100 = 7.69 \%.$
$(b)$ Molar mass of $C_2H_4O_2 = (2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \ g/mol$. Percentage of $C = (24 / 60) \times 100 = 40.0 \%$.
$(c)$ Molar mass of $CN_2H_2O = 12 + 14 + (2 \times 1) + 16 = 12 + 14 + 2 + 16 = 44 \ g/mol$. Percentage of $N = (14 / 44) \times 100 = 31.82 \%$.
$(d)$ Molar mass of $C_6H_6 = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \ g/mol$. Percentage of $H = (6 / 78) \times 100 = 7.69 \%.$
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