The molecular weight of calcium carbonate, CaCO3, is 100.09 g/mol. How many moles of calcium carbonate are in 63.8 g of calcium carbonate?
The Correct Answer and Explanation is:
To find how many moles of calcium carbonate (CaCO3) are in 63.8 grams of the substance, we will use the formula:moles=massmolar mass\text{moles} = \frac{\text{mass}}{\text{molar mass}}moles=molar massmass
Where:
- Mass is the given mass of calcium carbonate, 63.8 grams.
- Molar mass is the molecular weight of calcium carbonate, which is 100.09 g/mol.
Step-by-Step Calculation:
- Write the formula: moles of CaCO3=63.8 g100.09 g/mol\text{moles of CaCO3} = \frac{63.8 \text{ g}}{100.09 \text{ g/mol}}moles of CaCO3=100.09 g/mol63.8 g
- Perform the division: moles of CaCO3=0.638 mol\text{moles of CaCO3} = 0.638 \text{ mol}moles of CaCO3=0.638 mol
Final Answer:
There are 0.638 moles of calcium carbonate in 63.8 grams of calcium carbonate.
Explanation:
To understand this, remember that the molar mass of a substance tells you how much one mole of that substance weighs. In this case, 1 mole of calcium carbonate weighs 100.09 grams. Therefore, to find out how many moles are in 63.8 grams, we divide the mass of the sample by the molar mass. This gives us the number of moles of CaCO3 in the given sample.
This type of calculation is fundamental in stoichiometry, allowing you to convert between grams of a substance and moles, which is useful in chemical reactions where quantities of reactants and products are based on the mole concept.
