# Total no. of atoms present in 4.4g CO2 gas?

Apr 29, 2018

$6.0 \cdot {10}^{22} {\text{ atoms CO}}_{2}$

#### Explanation:

Molar mass of carbon is (12.011 "g")/"mol".

Molar mass of oxygen gas is (2 * 15.999 "g")/"mol" = (31.998 "g")/"mol".

So when you add them together to get the molar mass of ${\text{CO}}_{2}$ you get (44.009 "g")/"mol".

We also know that there are around $6.02 \cdot {10}^{23}$ atoms in a mole (Avogadro's number).

From this information, we can use dimensional analysis to solve for the number of atoms present in $4.4$ g ${\text{CO}}_{2}$ gas:
$4.4 {\text{g CO"_2 xx "1 mol CO"_2/(44.009 "g CO"_2) xx (6.02 * 10^23 " atoms CO"_2)/"1 mol CO}}_{2}$

$4.4 \cancel{{\text{g CO"_2) xx cancel("1 mol CO"_2)/(44.009 cancel("g CO"_2)) xx (6.02 * 10^23 " atoms CO"_2)/cancel("1 mol CO}}_{2}}$

$\approx 6.0187689 \ldots \cdot {10}^{22} {\text{ atoms CO}}_{2}$.

However, we can only have two significant figures because of the $4.4 \text{g}$ and also because it is multiplication, so the answer is actually:
$6.0 \cdot {10}^{22} {\text{ atoms CO}}_{2}$.

Hope this helps!