Question

How many grams are there in $1.5\,.\,{{10}^{26}}$ molecules of $C{{O}_{2}}$?

Answer 1

The amount of 1 mole which is the same as the gram atomic weight of a compound, always contains Avogadro number of molecules which is $6.022\times {{10}^{23}}$.

Complete step-by-step answer: We have been given the number of molecules as $1.5\,.\,{{10}^{26}}$, which are present in some unknown amount of carbon dioxide, $C{{O}_{2}}$. We have to determine this unknown mass (in grams) of this carbon dioxide.
As the molecules of $C{{O}_{2}}$, given are more in number than Avogadro number, which are $6.022\times {{10}^{23}}$molecules present in 1 mole of a compound. I inferred that the amount of carbon dioxide is more than 1 mole.
So, number of moles of$C{{O}_{2}}$, in $1.5\,.\,{{10}^{26}}$molecules of$C{{O}_{2}}$can be calculated by stoichiometric ratio factors as:
Moles of $C{{O}_{2}}$= $1.5\,.\,{{10}^{26}}$molecules $\times \dfrac{1\,\,mole\,\,C{{O}_{2}}}{6.022\times {{10}^{23\,}}\,molecules}$
Moles of $C{{O}_{2}}$= 2.491 $\times {{10}^{2}}$ moles
Now we have to calculate the value of these moles in grams. We know that 1 mole of any compound has mass equal to its molar mass. Here, 1 mole of carbon dioxide has molar mass of 44.01 g/mol. So, the mass in 2.491 $\times {{10}^{2}}$ moles of $C{{O}_{2}}$ will be:
Mass of $C{{O}_{2}}$= 2.491 $\times {{10}^{2}}$ moles $C{{O}_{2}}$ $\times \dfrac{44.01\,g}{1\,mole\,C{{O}_{2}}}$
Mass of $C{{O}_{2}}$= 109.63 g
Hence, the mass of $C{{O}_{2}}$in $1.5\,.\,{{10}^{26}}$molecules is 109.63 grams.

Note: The answer that we have calculated, can also be written as a round figure up to the last two significant figures. The answer rounded up to two significant figures will be 110 grams in $1.5\,.\,{{10}^{26}}$ molecules of carbon dioxide.