# How many moles are in 37.9 g of NaHCO3?

## (Useful data: Atomic masses: C = 12.01, O = 16.00, H = 1.01, Na = 22.99)

Apr 15, 2018

Number of moles = $\text{given mass"/"molar mass}$

$\implies \frac{37.9}{23 + 1 + 12 + 16 \left(3\right)}$

$\implies \frac{38}{84}$

$\implies 0.452 \text{ moles}$

Apr 15, 2018

$\approx 0.451 m o l$

#### Explanation:

Find molar mass of $N a H C {O}_{3}$ by adding the molar mass of all the individual atoms:
${M}_{r} = \left(22.99\right) + \left(1.01\right) + \left(12.01\right) + 3 \left(16.00\right) = 84.01 g m o {l}^{-} 1$

Use the formula:

$n = \frac{m}{M}$

$n$=number of moles
$m$=mass in grams of sample
$M$=molar mass of the compound

Hence:

$n = \frac{37.9}{84.01}$

$n \approx 0.451 m o l$