Answer : The mass percentage of carbon in sucrose is, 42.1 %
Explanation: Given,
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of O = 16 g/mole
First we have to calculate the molar mass of sucrose.
Molar mass of sucrose \((C_{12}H_{22}O_{11})\) = \(12(12)+22(1)+11(16)=342g/mole\)
Now we have to calculate the mass of carbon in 5 g of sucrose.
As we now that there are 12 number of carbon atoms, 22 number of hydrogen atoms and 11 number of oxygen atoms.
The mass of carbon = \(12\times 12=144g\)
As, 342 g of sucrose contains 144 g of carbon
So, 5 g of sucrose contains \(\frac{5}{342}\times 144=2.105g\) of carbon
Now we have to calculate the mass percentage of carbon in sucrose.
Formula used :
\(\%\text{ Mass percentage of carbon}=\frac{\text{Mass of carbon}}{\text{Mass of sucrose}}\times 100\)
Now put all the given values in this formula, we get:
\(\%\text{ Mass percentage of carbon}=\frac{2.105}{5}\times 100=42.1\%\)
Therefore, the mass percentage of carbon in sucrose is, 42.1 %
