# How many grams are there in 7.50 x 10^23 molecules of H_2SO_4?

May 14, 2018

Well, how many grams in $6.022 \times {10}^{23}$ $\text{sulfuric acid molecules}$?

#### Explanation:

It is a fact that such a quantity has a mass of $98.08 \cdot g$. Why? Because $6.022 \times {10}^{23}$ particles SPECIFIES a molar quantity. And we know (or can calculate) that sulfuric acid has a molar mass of $98.08 \cdot g$...

And so we take the quotient....

$\left(7.5 \times {10}^{23} \cdot \text{sulfuric acid particles")/(6.022xx10^23*"sulfuric acid particles} \cdot m o {l}^{-} 1\right) \times 98.08 \cdot g \cdot m o {l}^{-} 1 = 122 \cdot g$...

May 14, 2018

$122.5 g$

#### Explanation:

• 1st, we need to convert $7.50$x${10}^{23}$molecules into moles

$7.50$x${10}^{23}$molecules x $\frac{1 m o l}{6.02 x {10}^{23}}$ =$1.25 m o l$

• Now we convert $1.25 m o l$ to grams $\left(g\right)$

1.25cancel(mol x (98g)/(1cancel(mol) = $122.5 g$

• Note, when converting to grams, its gram formula mass / $1 m o l$. The formula mass is all the element's masses added together

${H}_{2}$ = $1 g$ x $2$ = $2 g$

$S$ = $32 g$

${O}_{4}$ = $16 g$ x $4$ = $64 g$

$2 g$ + $32 g$ + $64 g$ = $\frac{98 g}{1 m o l}$