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A 3.75-gram sample of limestone, [tex]CaCO_3[/tex], contains 1.50 g of calcium and 0.450 g of carbon. What is the percent oxygen in limestone?

Answer :

Final answer:

To calculate the percent oxygen in a 3.75-gram limestone (CaCO3) sample you subtract the weight of the calcium and carbon from the total weight to find the weight of the oxygen. Then, you divide the weight of the oxygen by the total weight and multiply by 100 to get the percentage. In this case, the percent oxygen in the limestone is 48%.

Explanation:

In order to determine the percent oxygen in limestone, we first have to identify the total weight of the limestone sample which is 3.75 g. We know 1.50 g is calcium and 0.450 g is carbon. So, if we subtract the weight of calcium and carbon from the total weight of the sample, we get the weight of oxygen.

In numeric terms: 3.75 g (total weight) - 1.50 g (calcium) - 0.450 g (carbon) = 1.80 g of oxygen in the limestone.

To find the percent oxygen, we divide the weight of oxygen by the total weight and multiply by 100. This gives us the formula: (1.80 g / 3.75 g) * 100 = 48%.

Therefore, the percent oxygen in the limestone sample is 48%.

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Rewritten by : Barada

Compute the mass of O in the given sample of CaCO₃ :

3.75 g - 1.50 g - 0.450 g = 1.80 g

This mass of oxygen makes up

(1.80 g) / (3.75 g) = 0.48 = 48%

of the sample's mass.