# How many grams of glucose are in 6.63 xx 10^23 molecules of glucose, C_6H_12O_6?

Feb 17, 2017

198 g of Glucose

#### Explanation:

We have to convert $6.63 \cdot {10}^{23}$ molecules of glucose into moles first.

To do this, we have to divide the number of molecules of glucose by Avogadro's Constant, $6.023 \cdot {10}^{23}$, which is the number of molecules of any substance in a mole of that substance.

(6.63*10^23 "molecules")/(6.023 * 10^23 "molecules/mol") = "1.101 mol" of glucose.

Then we can use the moles equation, Moles $= \frac{M a s s}{R M M}$

"1.101 mol" = (Mass)/([(6*12.01)+(12*1.008)+(6*16.00)] "g/mol")

$\text{1.101 mol" = (Mass)/"180.16 g/mol}$

$M a s s = \text{1.101 mol" * "180.16 g/mol}$

$M a s s = \text{198 g}$