what is the mass percent of carbon in propanol, c3h8o? report your answer to two decimal places. question 5…

Explanation:

Step1: Calculate molar mass of propanol ($C_3H_8O$)

The molar mass of $C = 12.01\ g/mol$, $H= 1.01\ g/mol$, $O = 16.00\ g/mol$. $M_{C_3H_8O}=3\times12.01 + 8\times1.01+16.00=36.03 + 8.08+16.00 = 60.11\ g/mol$

Step2: Calculate mass of carbon in propanol

There are 3 carbon atoms in $C_3H_8O$, so mass of carbon $m_C=3\times12.01 = 36.03\ g/mol$

Step3: Calculate mass - percent of carbon

Mass - percent of carbon $=\frac{m_C}{M_{C_3H_8O}}\times100%=\frac{36.03}{60.11}\times100%\approx59.94%$

Step4: Calculate molar mass of glucose ($C_6H_{12}O_6$)

$M_{C_6H_{12}O_6}=6\times12.01+12\times1.01 + 6\times16.00=72.06+12.12 + 96.00=180.18\ g/mol$

Step5: Calculate mass - fraction of carbon in glucose

There are 6 carbon atoms in $C_6H_{12}O_6$, mass of carbon in glucose $m = 6\times12.01=72.06\ g/mol$. Mass - fraction of carbon $w=\frac{72.06}{180.18}$

Step6: Calculate mass of carbon in 8.78 g of glucose

Mass of carbon $m_{carbon}=8.78\times\frac{72.06}{180.18}\approx3.51\ g$

Answer:

59.94% 3.51 g