176k views
4 votes
The density of calcium carbonate is 2.710 g/cm³. How many oxygen atoms are contained within 10.00 cm³ sample of calcium carbonate?

1 Answer

1 vote

Answer:

Step-by-step explanation:

The formula for calcium carbonate is CaCO3, which consists of one calcium atom, one carbon atom, and three oxygen atoms.

We can use the density and molar mass of calcium carbonate to find the number of moles of the compound in a 10.00 cm³ sample:

mass = density x volume

mass = 2.710 g/cm³ x 10.00 cm³

mass = 27.10 g

The molar mass of CaCO3 is the sum of the atomic masses of its elements, which are:

Calcium (Ca): 40.08 g/mol

Carbon (C): 12.01 g/mol

Oxygen (O): 16.00 g/mol (x 3 atoms)

Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (16.00 g/mol x 3) = 100.09 g/mol

Number of moles of CaCO3 = mass / molar mass = 27.10 g / 100.09 g/mol = 0.2703 mol

Since there are three oxygen atoms in one molecule of CaCO3, the number of oxygen atoms in the sample is:

Number of oxygen atoms = 0.2703 mol x 3 = 0.8109 mol of O atoms

To convert this to the number of oxygen atoms, we use Avogadro's constant, which is 6.022 × 10^23 particles per mole:

Number of oxygen atoms = 0.8109 mol x 6.022 × 10^23 atoms/mol ≈ 4.882 × 10^23 oxygen atoms

Therefore, there are approximately 4.882 × 10^23 oxygen atoms in a 10.00 cm³ sample of calcium carbonate.

User Lily Ballard
by
7.4k points