Answer
a) $288.5\ g/mol$
b) $C_{19}H_{28}O_2$
Work Step by Step
Atomic weights: C - 12.011, H - 1.008, O - 15.999
a) Calculate the percent by mass of O:
$\%m_O=100.00-79.12-9.79=11.09\%$
Given that there are two atoms of O per molecule, calculate the mass of oxygen per mol of this molecule:
$M_O=2\cdot 15.999=31.998\ g/mol$
Calculate the molar mass of the molecule:
$\%m_O=\frac{M_O}{M}\cdot 100.00\%$
$M=31.998\ g/mol\cdot\frac{100.00}{11.09}$
$M=288.5\ g/mol$
b) Calculate the mass of carbon and hydrogen atoms in 1 mol of this compound:
$m_C=288.5\ g/mol\cdot\frac{79.12}{100.00}=228.3\ g$
$m_H=288.5\ g/mol\cdot\frac{9.79}{100.00}=28.2\ g$
Calculate the number of moles of carbon and hydrogen atoms in 1 mol of this compound:
$n=\frac mM$
$n_C=\frac{228.3\ g}{12.011\ g/mol}\approx 19$
$n_H=\frac{28.2\ g}{1.008\ g/mol}\approx 28$
In 1 mol of this compound, there are 19 mols of carbon atoms and 28 mols of hydrogen atoms, so these are also the number of atoms of each of these elements in the molecule. Also, remember that there are 2 oxygen atoms per molecule, so the molecular formula is:
$C_{19}H_{28}O_2$
