Question

An organic compound is composed of 38.7% C, 9.70% H, 51.6% O. The compound has a molecular formula mass of 62.0g/mol.

Answer 1

B)C2H6O2

First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g. 
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.

Answer 2

Answer:

B) $C_{2}H_{6}O_{2}$

Explanation:

First you should add up the three percentages:

$38.7+9.70+51.6 = 100$

Then you take that base of 100g to calculate the mass of each element:

$C=38.7g$

$H=9.70g$

$O=51.6g$

Then you find the number of moles of each element using the molar mass of each one:

- For C:

$38.7gC*\frac{1molC}{12gC}=3.22molesC$

- For H:

$9.70gH*\frac{1molH}{1gH}=9.70molesH$

- For O:

$51.6gO*\frac{1molO}{16gO}=3.22molesO$

Then, you find the minimum number of moles and divide each one by it:

- For C:

$\frac{3.22}{3.22}=1$

- For H:

$\frac{9.70}{3.22}=3$

- For O:

$\frac{3.22}{3.22}=1$

You have find the proportion of each element in the compound that is $CH_{3}O$, but you should find the molecular formula taking in account the molar mass of the compound, so:

$(2.molarmassC)+(6.molarmassH)+(2.molarmassO)=62.0\frac{g}{mol}$

$(2*12\frac{g}{mol})+(6*1\frac{g}{mol})+(2*16\frac{g}{mol})=62.0\frac{g}{mol}$

So the molecular formula is $C_{2}H_{6}O_{2}$