# How do you calculate the formula mass for each of the following compounds? (a) C_6H_12O_6 (b) Fe(NO_3)_3?

Dec 27, 2016

a.) 180.156 g/mol
b.) 241.88 g/mol

#### Explanation:

$\textcolor{m a \ge n t a}{\text{We will use the formula below to answer each question:}}$

atomic weight of element $\times$number of atoms given by subscript $=$ formula mass

First, you want to have a periodic table available so you can determine the atomic weight of C, H, and O:

• C has an atomic weight of 12.01 g/mol
• H has an atomic weight of 1.008 g/mol
• O has an atomic weight of 16.00 g/mol

From the chemical formula, we have 6 carbon atoms. Therefore, the atomic mass has to be multiplied by 6 to obtain an atomic weight of 72.06 g/mol.

There are 12 hydrogen atoms, so the atomic mass has to be multiplied by 12, which gives you a value of 12.096 g/mol

Next, you have 6 oxygen atoms, so you would multiply the atomic mass of O by 6 to obtain an atomic weight of 96.00 g/mol

Now you want to add the mass of each atom together to obtain the formula mass of the entire compound:

$\textcolor{red}{\text{72.06 g/mol + 12.096 g/mol + 96.00 g/mol = 180.156 g/mol}}$

For $F e {\left(N {O}_{3}\right)}_{3}$, we do the same thing that was done above.

• Fe has an atomic weight of 55.85 g/mol
• N has an atomic weight of 14.01 g/mol
• O has an atomic weight of 16.00 g/mol

We have 1 iron atom. So the atomic mass is just 55.85 g/mol.

There are 3 nitrogen atoms, so the atomic mass has to be multiplied by 3, which gives you a value of 42.03 g/mol

Next, you have 9 oxygen atoms, so you would multiply the atomic mass of O by 9 to obtain an atomic weight of 144.00 g/mol

$\textcolor{p u r p \le}{\text{55.85 g/mol + 42.03 g/mol + 144.00 g/mol = 241.88 g/mol}}$