Answer
Given amount of ethanol contains more carbon than given amount of glucose. 5.85 g of ethanol $C_{2}H_{6}O$ contains 3.05 g C that is greater than 2.40 g C contained in 6.01 g of glucose $C_{6}H_{12}O_{6}$.
Work Step by Step
Strategy: To find which of the given amounts of compaunds contains more C : 1- find the Molar mass of each compound 2- Calculate the percentage of C in each 3 - Calculate the amount of C in the given amount of each compound 4 - Compare the amounts
Glucose $C_{6}H_{12}O_{6}$.
Molar mass $C_{6}H_{12}O_{6}$ = 6 x 12.0 + 12 x 1.00 + 6 x 16
Molar mass $C_{6}H_{12}O_{6}$ = 180 g/mol
Mass percentage C ( in $C_{6}H_{12}O_{6}$) = $\frac{mass_(6 mole C)}{Molar mass_(C_{6}H_{12}O_{6})}$$\times$ 100%
Mass percentage C ( in $C_{6}H_{12}O_{6}$) = $\frac{mass_(6\times12.0)}{180}$$\times$ 100%
Mass percentage C ( in $C_{6}H_{12}O_{6}$) = 40%
The mass of C in 6.01 g in $C_{6}H_{12}O_{6}$ is 40% of 6.01:
40%$\times$6.01g = $\frac{40}{100}\times6.01$
The mass of C in 6.01 g in $C_{6}H_{12}O_{6}$ is 2.40 g.
Ethanol $C_{2}H_{6}O$.
Molar mass $C_{2}H_{6}O$ = 2 x 12.0 + 6 x 1.00 + 1 x 16
Molar mass $C_{2}H_{6}O$ = 46 g/mol
Mass percentage C ( in $C_{2}H_{6}O$) = $\frac{mass_(2 mole C)}{Molar mass_(C_{2}H_{6}O)}$$\times$ 100%
Mass percentage C ( in $C_{2}H_{6}O$) = $\frac{mass_(2\times12.0)}{180}$$\times$ 100%
Mass percentage C ( in $C_{2}H_{6}O$) = 52.2%
The mass of C in 5.85 g in $C_{2}H_{6}O$ is 52.2% of 5.85g :
52.2%$\times$5.85g = $\frac{52.2}{100}\times5.85g$
The mass of C in 5.85 g $C_{2}H_{6}O$ is 3.05 g.
Given amount of ethanol contains more carbon than given amount of glucose. 5.85 g of ethanol $C_{2}H_{6}O$ contains 3.05 g C that is greater than 2.40 g C contained in 6.01 g of glucose $C_{6}H_{12}O_{6}$.
