# What is the number of moles in .197 g As_2O_3?

#### Answer:

Do you mean 0.197 g?
Then,
number of moles = 0.197 $\div$molar mass

#### Explanation:

Do you mean 0.197 g?

Assume that:
n = number of moles
m = mass of substance
M = molar mass (equivalent to atomic weight on periodic table)

n = m $\div$M

0.197 g (mass) of $A {s}_{2} {O}_{3}$ is provided for you.

The next step is to find the molar mass (M). First thing first, you should know what elements are involved. As you can see, arsenic and oxygen are involved.

If you refer to your periodic table, the molar mass of arsenic is 74.9 g/mol and oxygen is 16.0 g/mol.

Since there are 2 arsenic and 3 oxygen involved, the molar mass is $\left[2 \times 74.9 + 3 \times 16.0\right]$ = 197.8 g/mol.

You've now found the molar mass of $A {s}_{2} {O}_{3}$.

The final step is to find out the number of moles (n). Looking back at the $n = m \div M$ formula, the n is now the unknown.

The number of moles present in $A {s}_{2} {O}_{3}$ is:
n = 0.197 moles $\div$197.8 g/mol = 0.000996 moles. Note that this answer is rounded to 3 significant figures.