# Question a95a2

##### 1 Answer
Mar 3, 2015

!!LONG ANSWER !!

Let's start with the number of nickel atoms, since it's the easiest one to calculate.

Before determining how many atoms of nickel you have in 1 gram, you must first determine how many moles of nickel you have. This is done by using its molar mass, 58.7 g/mol.

$\text{1 g" * "1 mole"/"58.7 g" = "0.0170 moles Ni}$

Use Avodagro's number - $6.022 \cdot {10}^{23}$ atoms in 1 mole - to determine how many atoms of nickel you have

$\text{0.0170 moles" * (6.022 * 10^(23)"atoms")/"1 mole" = 1.02 * 10^(22)"atoms of Ni}$

Now for the lithium aluminium hydride (LAH), or $L i A l {H}_{4}$. Notice that 1 molecule of LAH has 1 atom of lithium, 1 of aluminium, and 4 of hydrogen. Determine how many molecules of LAH you have first, then go to its component atoms.

$\text{0.10 g" * "1 mole"/"38.0 g" = "0.00263 moles LAH}$

$\text{0.00263 moles" * (6.022 * 10^(23)"molecules")/"1 mole" = 1.6 * 10^(21)"molecules of LAH}$

The number of atoms of Li, Al, and H will be

$1.6 \cdot {10}^{21} \text{molecules LAH" * "1 atom Li"/"1 molecule LAH" = 1.6 * 10^(21)"atoms Li}$

$1.6 \cdot {10}^{21} \text{molecules LAH" * "1 atom Al"/"1 molecule LAH" = 1.6 * 10^(21)"atoms Al}$

$1.6 \cdot {10}^{21} \text{molecules LAH" * "4 atoms H"/"1 molecule LAH" = 6.4 * 10^(21)"atoms H}$

Now for the oxygen. Gases are a little trickier because you need values for temperature and pressure to determine the number of moles. Since no mention of particular conditions was made, I'll assume STP - 273.15 K and 1 atm.

The great part about STP is that 1 mole of any ideal gas occupies exactly 22.4 L - this is called the molar volume of a gas. So if you know the volume, you can determine how many moles of oxygen you have.

Remember that air is only 20.9% oxygen, so the volume of oxygen gas will be

$\text{1 mL air" * "20.9 mL oxygen"/"100 mL air" = "0.209 mL }$ ${O}_{2}$

Another thing to look out for - notice that you've got diatomic oxygen, or ${O}_{2}$. This means that the number of oxygen atoms will be twice the number of ${O}_{2}$ molecules. So,

n = V/V_("molar") = (0.209 * 10^(-3)"L")/"22.4 L" = 9.3 * 10^(-6)"moles "# ${O}_{2}$

$9.3 \cdot {10}^{- 6} \text{moles" * (6.022 * 10^(23)"molecules" O_2)/("1 mole"O_2) = 5.6 * 10^(18)"molecules oxygen}$

Therefore, the number of oxygen atoms will be

$5.6 \cdot {10}^{18} \text{molecules oxygen" * "2 oxygen atoms"/"1 oxygen molecule" = 1.1 * 10^(19)"atoms of oxygen}$

You can use the same strategy to calculate how many atoms of nitrogen you have. Keep in mind that it too is a diatomic molecule, ${N}_{2}$, and that it's percentage in air's composition is 78.1%.