How many moles are in 37.9 g of NaHCO3?

Question

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

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Answer 1 · 2018-04-15T10:07:21

Number of moles = $"given mass"/"molar mass"$

$=> 37.9/(23+1+12+16(3))$

$=> 38/84$

$=> 0.452 " moles"$

Answer 2 · 2018-04-15T10:10:19

$approx 0.451 mol$

Find molar mass of $NaHCO_3$ by adding the molar mass of all the individual atoms:
$M_r=(22.99)+(1.01)+(12.01)+3(16.00)=84.01 gmol^-1$

Use the formula:

$n=(m)/M$

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

Hence:

$n=(37.9)/84.01$

$n approx 0.451 mol$