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 [tex](C_{12}H_{22}O_{11})[/tex] = [tex]12(12)+22(1)+11(16)=342g/mole[/tex]
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 = [tex]12\times 12=144g[/tex]
As, 342 g of sucrose contains 144 g of carbon
So, 5 g of sucrose contains [tex]\frac{5}{342}\times 144=2.105g[/tex] of carbon
Now we have to calculate the mass percentage of carbon in sucrose.
Formula used :
[tex]\%\text{ Mass percentage of carbon}=\frac{\text{Mass of carbon}}{\text{Mass of sucrose}}\times 100[/tex]
Now put all the given values in this formula, we get:
[tex]\%\text{ Mass percentage of carbon}=\frac{2.105}{5}\times 100=42.1\%[/tex]
Therefore, the mass percentage of carbon in sucrose is, 42.1 %
