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Question: Total no. of atoms present in \( 4.4 \cdot g \cdot C{O_2}\) gas?...

Total no. of atoms present in 4.4gCO2 4.4 \cdot g \cdot C{O_2} gas?

Explanation

Solution

The amount of substance present in the one mole is equal to its molecular mass. The same concept can be utilised in this problem too. One mole of carbon dioxide will have 44g44g of carbon dioxide and one mole of substance contains 6.022×10236.022 \times {10^{23}} particles. Here, the question is about the total number of atoms, not the number of molecules or particles of carbon dioxide.

Complete answer:
Moles are the unit of measurement of the amount of a substance as per International System of Units. It is defined as exactly Avogadro’s number i.e. 6.022×10236.022 \times {10^{23}}. One mole of carbon dioxide contains 44g44g of carbon dioxide.
Let us calculate the molar mass of carbon dioxide –
Molar mass of carbon dioxide is equal to the molar mass of carbon multiplied by the twice of molar mass of oxygen, i.e.
\eqalign{ & {M_{C{O_2}}} = {M_C} + 2({M_O}) \cr & {M_{C{O_2}}} = 12 + 2(16) \cr & {M_{C{O_2}}} = 44g \cr}
Now, 44g44g of carbon dioxide will have 6.022×10236.022 \times {10^{23}} molecules and each molecule contains three atoms, one carbon atom and two oxygen atoms.
From this information, we can use dimensional analysis to solve for the number of atoms present in 4.4gCO2 4.4 \cdot g \cdot C{O_2} gas:
\eqalign{ & 4.4 \cdot g \cdot C{O_2} \times \dfrac{{1 \cdot mol \cdot C{O_2}}}{{44 \cdot g \cdot C{O_2}}} \times \dfrac{{6.022 \times {{10}^{23}} \cdot molecules \cdot C{O_2}}}{{1 \cdot mol \cdot C{O_2}}} \cr & 4.4g{\text{ }}C{O_2} \times \dfrac{{1 \cdot mol \cdot C{O_2}}}{{44 \cdot g \cdot C{O_2}}} \times \dfrac{{6.022 \times {{10}^{23}} \cdot molecules \cdot C{O_2}}}{{1 \cdot mol \cdot C{O_2}}} \cr & = 6.022 \times {10^{22}} \cdot molecules \cdot C{O_2} \cr}
However, we can only have two significant figures because of the 4.4g   4.4g\; and also because it is multiplication, so the answer is actually: 6.0×1022moleculesCO2 6.0 \times {10^{22}} \cdot molecules \cdot C{O_2} .
Now, each molecule have three atoms in each, so the number of total atoms will be - 3×(6.0×1022CO2)=18×1022atoms3 \times (6.0 \times {10^{22}}C{O_2}) = 18 \times {10^{22}} \cdot atoms .
Hence, the total number of atoms present in 4.4gCO2 4.4 \cdot g \cdot C{O_2} gas is 18×1022atoms18 \times {10^{22}} \cdot atoms .

Note:
It was confirmed that there are two oxygen atoms in one molecule of carbon dioxide. This is known as atomicity. The number of atoms that make up a molecule. It is determined by the ratio of molecular weight to atomic weight.