How many grams are there in $7.50 x 10^{23}$ $7.50 x 10^23$ molecules of $H_{2} S O_{4}$ $H_2SO_4$ ?

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How many grams are there in $7.50 x 10^{23}$ $7.50 x 10^23$ molecules of $H_{2} S O_{4}$ $H_2SO_4$ ?

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Answer 1 · 2018-05-14T07:00:34

Well, how many grams in $6.022 \times 10^{23}$ $6.022xx10^23$ $sulfuric acid molecules$ $"sulfuric acid molecules"$ ?

Explanation:

It is a fact that such a quantity has a mass of $98.08 \cdot g$ $98.08*g$ . Why? Because $6.022 \times 10^{23}$ $6.022xx10^23$ particles SPECIFIES a molar quantity. And we know (or can calculate) that sulfuric acid has a molar mass of $98.08 \cdot g$ $98.08*g$ ...

And so we take the quotient....

$\frac{7.5 \times 10^{23} \cdot sulfuric acid particles}{6.022 \times 10^{23} \cdot sulfuric acid particles \cdot m o l^{- 1}} \times 98.08 \cdot g \cdot m o l^{- 1} = 122 \cdot g$ $(7.5xx10^23*"sulfuric acid particles")/(6.022xx10^23*"sulfuric acid particles"*mol^-1)xx98.08*g*mol^-1=122*g$ ...

Answer 2 · 2018-05-14T07:03:26

$122.5 g$ $122.5g$

Explanation:

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

$7.50$ $7.50$ x $10^{23}$ $10^23$ molecules x $\frac{1 m o l}{6.02 x 10^{23}}$ $(1mol)/(6.02x10^23)$ = $1.25 m o l$ $1.25mol$

Now we convert $1.25 m o l$ $1.25mol$ to grams $(g)$ $(g)$

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

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

$H_2$ = $1g$ x $2$ = $2g$

$S$ = $32g$

$O_4$ = $16g$ x $4$ = $64g$

$2g$ + $32g$ + $64g$ = $(98g)/(1mol)$

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How many grams are there in $7.50 x 10^{23}$ $7.50 x 10^23$ molecules of $H_{2} S O_{4}$ $H_2SO_4$ ?

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