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Answer 1 · 2017-10-03T03:09:49

$1.204* 10^23$

![https://uachemistry13.wordpress.com/2013/07/19/the-point-of-the-mole/]( useruploads.socratic.org )

$"4.032 g"$ divided by the molar mass of $"MgO"$ , $"40.3044 g/mol"$ , gives you moles

$"4.032 g"/"40.3044 g/mol" = "0.10004 moles"$

Then multiply this by $6.02 * 10^23$ to get the number of formula units, which would be

$0.10004 cancel("moles") * (6.02* 10^23"f. units")/(1cancel("mole")) = 0.60224 * 10^(23)$ $"f. units"$ .

Since one formula unit of $"MgO"$ contains $2$ atoms--one of $"Mg"$ and one of $"O"$ --multiply this by $2$ to get

$0.60224 * 10^(23)cancel("f. units") * "2 atoms"/(1cancel("f. unit")) = 1.204 * 10^(23)$ $"atoms"$