Question

empirical formula & molecular formula ws
remember: percent to mass, mass to moles, divide by the smallest, & (if necessary) multiply til whole

1. a compound is found to have (by mass) 48.38% carbon, 8.12% hydrogen and the rest oxygen. what is its empirical formula?
2. a compound is found to have 46.67% nitrogen, 6.70% hydrogen, 19.98% carbon and 26.65% oxygen. what is its empirical formula?
3. a compound is known to have an empirical formula of ch and a molar mass of 78.11 g/mol. what is its molecular formula?
4. another compound, also with an empirical formula if ch is found to have a molar mass of 26.04 g/mol. what is its molecular formula?
5. a compound is found to have 1.121 g nitrogen, 0.161 g hydrogen, 0.480 g carbon and 0.640 g oxygen. what is its empirical formula? (note that masses are given, not percentages.)

ws: molecular and empirical formulas
write the molecular formulas of the following compounds:

1. a compound with an empirical formula of c₂oh₄ and a molar mass of 88 grams per mole.
2. a compound with an empirical formula of c₂h₂o and a molar mass of 136 grams per mole.
3. a compound with an empirical formula of cfbro and a molar mass of 254.7 grams per mole.
4. a compound with an empirical formula of c₂h₅n and a molar mass of 46 grams per mole.

determination of hydrate number, n, for a hydrated ionic compound
a 5.00 g sample of hydrated barium chloride, bacl₂·nh₂o, is heated to drive off the water. after heating, 4.25 g of anhydrous barium chloride, bacl₂, remains. what is the value of n in the hydrates formula?

Explanation:

Let's solve the first problem (finding the empirical formula of a compound with 48.38% C, 8.12% H, and the rest O) step by step.

Step 1: Assume 100g of the compound

If we assume we have 100g of the compound, then the masses of each element are:

• Carbon (C): \( 48.38 \, \text{g} \)
• Hydrogen (H): \( 8.12 \, \text{g} \)
• Oxygen (O): \( 100 - 48.38 - 8.12 = 43.5 \, \text{g} \)

Step 2: Convert masses to moles

Using the molar masses:

• Molar mass of C: \( 12.01 \, \text{g/mol} \)
• Molar mass of H: \( 1.008 \, \text{g/mol} \)
• Molar mass of O: \( 16.00 \, \text{g/mol} \)

Moles of C: \( \frac{48.38}{12.01} \approx 4.028 \, \text{mol} \)
Moles of H: \( \frac{8.12}{1.008} \approx 8.055 \, \text{mol} \)
Moles of O: \( \frac{43.5}{16.00} \approx 2.719 \, \text{mol} \)

Step 3: Divide by the smallest number of moles

The smallest number of moles is approximately 2.719 (from O).

Divide each by 2.719:

• C: \( \frac{4.028}{2.719} \approx 1.48 \approx 1.5 \) (we can multiply by 2 to get whole numbers)
• H: \( \frac{8.055}{2.719} \approx 2.96 \approx 3 \) (multiply by 2)
• O: \( \frac{2.719}{2.719} = 1 \) (multiply by 2)

After multiplying by 2:

• C: \( 1.5 \times 2 = 3 \)
• H: \( 3 \times 2 = 6 \)
• O: \( 1 \times 2 = 2 \)

Answer:

The empirical formula is \( \text{C}_3\text{H}_6\text{O}_2 \)