Question

How many grams of glucose are in $6.63 \times {10^{23}}$ molecules of glucose, ${C_6}{H_{12}}{O_6}$ ?

Answer 1

Mole is one of the units used to represent the amount of substance. The number of molecules or atoms or ions present in one mole is equal to $6.023 \times {10^{23}}$ . By determining the moles from the given molecules and substituting the obtained moles and molar mass in the below formula gives the mass of glucose in grams.
$n = \dfrac{m}{M}$
$n$ is number of moles of glucose
$m$ is mass of glucose
$M$ is the molar mass of glucose.

Complete Step By Step Answer:
Given that the number of molecules is $6.63 \times {10^{23}}$
Mole is the concept used to represent the small quantity of chemical substances. It is used to express the number of atoms or ions or molecules in one mole.
One mole is equal to $6.023 \times {10^{23}}$ , This number is known as Avogadro’s number.
Thus, the number of molecules given must be divided with Avogadro’s number to obtain the number of moles of glucose.
$\dfrac{{6.63 \times {{10}^{23}}molecules}}{{6.023 \times {{10}^{23}}molecules{{\left( {moles} \right)}^{ - 1}}}} = 1.1moles$
Glucose is a monosaccharide with the chemical formula of ${C_6}{H_{12}}{O_6}$ whose molar mass is $180gmo{l^{ - 1}}$
The number of moles is the ratio of the mass to molar mass of that substance.
$1.1moles = \dfrac{m}{{180gmo{l^{ - 1}}}}$
Further simplification, the mass will be $m = 198g$
Thus, $198g$ grams of glucose are in $6.63 \times {10^{23}}$ molecules of glucose, ${C_6}{H_{12}}{O_6}$ .

Note:
Based on the number of atoms in the glucose, the molar mass should be taken. The atoms molar mass must be taken based on the periodic table only. as the molar mass was taken in grams per mole, the moles must be in moles only, but not molecules. Thus, the conversion must be made from molecules to moles.