What is the mass of 1.56 × 10^21 atoms of magnesium in grams? I cant figure out how to do this
You will need Avogadro's number and the molecular mass of Mg from the periodic table. 1. Convert atoms to moles 2. Convert moles to grams The answer will provide you with the number of grams of Mg when the other units cancel out.
\[1.56 X 10^{21} atoms Mg \times \left( \frac{ 1 mol Mg }{ 6.022\times10^{23} atoms of Mg } \right) \times \left( \frac{ MM of Mg from the periodic table }{ 1 mol Mg } \right)\]
I know it got cut off at the end, but the second part just says to put the MM of Mg from the periodic table above 1 mol Mg and the atoms will cancel out, the mol Mg will cancel out and your answer will be in g of Mg. Hope it helps.
\[1\text{ mole} = 6.022*10^{23}\text{ particles}\]Here our particles are atoms, so we have \[1\text{ mole} = 6.022*10^{23}\text{ atoms}\]We also know that \(1\) mole of anything has the same mass (in grams) as the it does the mass of \(1\) particle in atomic mass units. That means \(1\) mole of Mg will have a mass of \(24.3050\) grams because that is the atomic mass of Mg. All we need to do is find out how many moles we have, and then multiply the number of moles by the mass of \(1\) mole. Finding the number of moles is easy: \[\frac{\text{# of atoms}}{\text{Avogadro's constant}} = \text{# of moles}\] we can also write that as \[\text{# of atoms} * \frac{1\text{ mole}}{6.022*10^{23}\text{ atoms}}\] and the whole enchilada will be: \[\text{mass of Mg in g} = \text{# of atoms of Mg} * \frac{1\text{ mole}}{6.022*10^{23}\text{ atoms}}*\frac{24.3050\text{ g}}{1\text{ mole}}\] notice that the units cancel out: \[\text{mass of Mg in g} = \text{# of }\cancel{\text{atoms}}\text{ of Mg} * \frac{1\cancel{\text{ mole}}}{6.022*10^{23}\cancel{\text{ atoms}}}*\frac{24.3050\text{ g}}{1\cancel{\text{ mole}}}\]leaving us with \[\text{mass of Mg in g} = \text{<some number> g}\] If they don't cancel out like that, it means we have blundered somewhere.
\[\text{<some number>} = 1.56*10^{21}*\frac{24.3050}{6.022*10^{23}}\]
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