# What mass of magnesium contains the same number of atoms as 8.0g of calcium?

Jan 15, 2018

$\text{4.9 g Mg}$

#### Explanation:

Molar Mass of Magnesium ("Mg"): " 24.31 g/mol"
Molar Mass of Calcium ("Ca"): " 40.08 g/mol"

Avogadro's constant $= 6.022 \cdot {10}^{23}$ particles of a substance per mol of that substance

$8.0 \text{g Ca" * (1 "mol Ca")/(40.08"g Ca")*(6.022*10^23"atoms of Ca")/(1 "mol Ca") = 1.2*10^23"atoms of Ca}$

The number of atoms of Mg should be the same as the number of atoms of Ca, so:

$1.2 \cdot {10}^{23} \text{atoms of Mg"*(1 "mol Mg")/(6.022*10^23"atoms of Mg")*(24.31"g Mg")/(1"mol Mg") = 4.9"g Mg}$

You can also eliminate the step of converting to atoms, as a mol of any substance contains $6.022 \cdot {10}^{23}$ particles of that substance