An organic compound has 42.1 % carbon,6.4 % h | vedclass

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(D) $1$. Calculate the percentage of oxygen: $100 - (42.1 + 6.4) = 51.5 \%$. $2$. Calculate the number of moles of each element in $100 \ g$ of the co

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$C_{12}H_{22}O_{11}$

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(D) $1$. Calculate the percentage of oxygen: $100 - (42.1 + 6.4) = 51.5 \%$. $2$. Calculate the number of moles of each element in $100 \ g$ of the compound: $n_C = \frac{42.1}{12} \approx 3.51$,$n_H = \frac{6.4}{1} = 6.4$,$n_O = \frac{51.5}{16} \approx 3.22$. $3$. Determine the empirical formula ratio: $C : H : O = \frac{3.51}{3.22} : \frac{6.4}{3.22} : \frac{3.22}{3.22} \approx 1.09 : 1.99 : 1 \approx 1.1 : 2 : 1$. $4$. Multiplying by $10$ gives $C_{11}H_{20}O_{10}$ (Empirical mass $\approx 312$). $5$. Given molecular weight is $342$. For $C_{12}H_{22}O_{11}$,molecular weight $= (12 	imes 12) + (22 	imes 1) + (11 	imes 16) = 144 + 22 + 176 = 342$. $6$. Thus,the molecular formula is $C_{12}H_{22}O_{11}$.

An organic compound has $42.1 \%$ carbon,$6.4 \%$ hydrogen and the remainder is oxygen. If its molecular weight is $342$,then its molecular formula is:

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