# Question #a1c12

Oct 10, 2015

The empirical formula is $\text{C"_4"H"_5"N"_2"O}$.

The molecular formula is $\text{C"_8"H"_10"N"_4"O"_2}$

#### Explanation:

Assume that you have $\text{195 g (1mol)}$ of the compound.

Determine the mass of each element in the compound by multiplying its percentage times its molar mass $\left(\text{195 g}\right)$.

$0.495 \times 195 \text{g"=96.5 "g C}$
$0.0515 \times 195 \text{g"=10.0 "g H}$
$0.289 \times 195 \text{g"=56.4 "g N}$
$0.165 \times 195 \text{g"=32.2 "g O}$

Determine the Molecular Formula
Determine the number of moles of each element using its calculated mass and its molar mass (atomic weight on the periodic table in grams/mole).

For example, the molar mass of $\text{C}$ is 12.0107 g/mol, which means that 1 mol of C has a mass of 12.0107 g. We can convert that into two conversion factors: $\left(12.0107 \text{g C")/(1"mol C}\right)$ and $\left(1 \text{mol C")/(12.0107"g C}\right)$. We will use the second conversion factor, so we will essentially be dividing the mass of the element in the compound by its molar mass.

$96.5 \cancel{\text{g C"xx(1"mol C")/(12.0107 cancel"g C")="8.04 mol C}}$$\approx \text{8 mol C}$

$10.0 \cancel{\text{g H"xx(1"mol H")/(1.00794 cancel"g H")="9.92 mol H}}$$\approx \text{10 mol H}$

$56.4 \cancel{\text{g N"xx(1"mol N")/(14.0067 cancel"g N")="4.03 mol H}}$$\approx \text{4 mol N}$

$32.2 \cancel{\text{g O"xx(1"mol O")/(15.999 cancel"g O")="2.01 mol O}}$$\approx \text{2 mol O}$

The molecular formula is $\text{C"_8"H"_10"N"_4"O"_2}$.

Emperical Formula

The empirical formula for a compound represents the lowest whole number ratio of elements in the compound. If we look at the molecular formula, $\text{C"_8"H"_10"N"_4"O"_2}$, we see that the subscripts have a common factor of $2$, which we can factor out of the molecular formula.

Therefore, the empirical formula is $\text{C"_4"H"_5"N"_2"O}$.