QuestionAugust 20, 2025

How many formula units are present in 872 grams of lead (IV)carbonate?

How many formula units are present in 872 grams of lead (IV)carbonate?
How many formula units are present in 872 grams of lead (IV)carbonate?

Solution
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Answer

1.58 \times 10^{24} formula units Explanation 1. Calculate Molar Mass of Lead (IV) Carbonate Lead (IV) carbonate is Pb(CO_3)_2. Molar mass = 207.2 + 2 \times (12.01 + 3 \times 16.00) = 331.2 \, \text{g/mol}. 2. Calculate Moles of Lead (IV) Carbonate Use the formula: \text{moles} = \frac{\text{mass}}{\text{molar mass}}. Moles = \frac{872}{331.2} \approx 2.63 moles. 3. Calculate Formula Units Use Avogadro's number: 6.022 \times 10^{23} formula units/mole. Formula units = 2.63 \times 6.022 \times 10^{23} \approx 1.58 \times 10^{24}.

Explanation

1. Calculate Molar Mass of Lead (IV) Carbonate<br /> Lead (IV) carbonate is $Pb(CO_3)_2$. Molar mass = $207.2 + 2 \times (12.01 + 3 \times 16.00) = 331.2 \, \text{g/mol}$.<br />2. Calculate Moles of Lead (IV) Carbonate<br /> Use the formula: $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$. Moles = $\frac{872}{331.2} \approx 2.63$ moles.<br />3. Calculate Formula Units<br /> Use Avogadro's number: $6.022 \times 10^{23}$ formula units/mole. Formula units = $2.63 \times 6.022 \times 10^{23} \approx 1.58 \times 10^{24}$.
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