Question

practice: determining formulas part 1: determine the empirical formula (aka, the lowest whole number ratio) for each compound below. show all work! 1. 13% magnesium and 87% bromine 2. 75% carbon and 25% hydrogen 3. 32.38% sodium, 22.65% sulfur, and 44.99% oxygen 4. 22.1% aluminum, 25.4% phosphorus, 52.5% oxygen 5. 63.5% silver, 8.2% nitrogen, 28.3% oxygen part 2: determine the molecular formula for each compound given the provided information. show all work! 6. consider the empirical formula you determined in #5. if that compound has a molar mass of 170 g/mol, what is the molecular formula? 7. if the empirical formula is ch and the molar mass is 78 g/mol, what is the molecular formula?

Explanation:

Step1: Calculate moles for Part 1 problems

Assume 100g of the compound. Then the mass of each element is equal to its percentage. Use the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass, and $M$ is the molar - mass. For example, for magnesium ($M_{Mg}=24.31g/mol$) and bromine ($M_{Br}=79.90g/mol$) in problem 1:
For magnesium: $n_{Mg}=\frac{13g}{24.31g/mol}\approx0.535mol$
For bromine: $n_{Br}=\frac{87g}{79.90g/mol}\approx1.09mol$
Then find the mole - ratio by dividing each number of moles by the smallest number of moles among them. $\frac{n_{Br}}{n_{Mg}}=\frac{1.09mol}{0.535mol}\approx2$, so the empirical formula is $MgBr_{2}$.
Repeat this process for problems 2 - 5.
For problem 2:
Carbon ($M_{C}=12.01g/mol$), $n_{C}=\frac{75g}{12.01g/mol}\approx6.24mol$
Hydrogen ($M_{H}=1.01g/mol$), $n_{H}=\frac{25g}{1.01g/mol}\approx24.75mol$
$\frac{n_{H}}{n_{C}}=\frac{24.75mol}{6.24mol}\approx4$, empirical formula is $CH_{4}$.
For problem 3:
Sodium ($M_{Na}=22.99g/mol$), $n_{Na}=\frac{32.38g}{22.99g/mol}\approx1.41mol$
Sulfur ($M_{S}=32.07g/mol$), $n_{S}=\frac{22.65g}{32.07g/mol}\approx0.706mol$
Oxygen ($M_{O}=16.00g/mol$), $n_{O}=\frac{44.99g}{16.00g/mol}\approx2.81mol$
Dividing by 0.706mol: $n_{Na}\approx2$, $n_{S}=1$, $n_{O}\approx4$, empirical formula is $Na_{2}SO_{4}$.
For problem 4:
Aluminum ($M_{Al}=26.98g/mol$), $n_{Al}=\frac{22.1g}{26.98g/mol}\approx0.82mol$
Phosphorus ($M_{P}=30.97g/mol$), $n_{P}=\frac{25.4g}{30.97g/mol}\approx0.82mol$
Oxygen ($M_{O}=16.00g/mol$), $n_{O}=\frac{52.5g}{16.00g/mol}\approx3.28mol$
Dividing by 0.82mol: $n_{Al}=1$, $n_{P}=1$, $n_{O}=4$, empirical formula is $AlPO_{4}$.
For problem 5:
Silver ($M_{Ag}=107.87g/mol$), $n_{Ag}=\frac{63.5g}{107.87g/mol}\approx0.59mol$
Nitrogen ($M_{N}=14.01g/mol$), $n_{N}=\frac{8.2g}{14.01g/mol}\approx0.59mol$
Oxygen ($M_{O}=16.00g/mol$), $n_{O}=\frac{28.3g}{16.00g/mol}\approx1.77mol$
Dividing by 0.59mol: $n_{Ag}=1$, $n_{N}=1$, $n_{O}=3$, empirical formula is $AgNO_{3}$.

Step2: Calculate the molecular formula for Part 2 problems

For problem 6, the empirical formula is $AgNO_{3}$ with an empirical - formula mass $M_{empirical}=107.87g/mol + 14.01g/mol+3\times16.00g/mol=169.88g/mol\approx170g/mol$. Since the molar mass of the compound is 170g/mol and the empirical - formula mass is approximately 170g/mol, the molecular formula is $AgNO_{3}$.
For problem 7, the empirical formula is $CH$ with an empirical - formula mass $M_{empirical}=12.01g/mol + 1.01g/mol = 13.02g/mol$.
Let $n=\frac{M_{molar}}{M_{empirical}}=\frac{78g/mol}{13.02g/mol}\approx6$.
The molecular formula is $C_{6}H_{6}$.

Answer:

1. $MgBr_{2}$
2. $CH_{4}$
3. $Na_{2}SO_{4}$
4. $AlPO_{4}$
5. $AgNO_{3}$
6. $AgNO_{3}$
7. $C_{6}H_{6}$