# The mass fraction of impurities in limestone is 6.25%. Calculate the mass of the basic substance (calcium carbonate)

**The mass fraction of impurities in limestone is 6.25%. Calculate the mass of the basic substance (calcium carbonate) contained in 1.5 tons of natural limestone.**

Given: ω (impurities) = 6.25%, m (sample) = 1.5 tons.

Find: ω (basic substance) -?

Decision.

1. Let’s calculate the mass of impurities in a sample of natural limestone:

m (impurities) = (ω (impurities) • m (sample)): 100% = 6.25% • 1.5 t: 100% = 0.09375 t.

2. Calculate the mass of the main substance:

m (basic substance) = m (sample) – m (impurities) = 1.5 t – 0.09375 g = 1.40625 t.

2. Let’s calculate the mass fraction of the main substance:

ω (basic substance) = m (basic substance) / m (sample) = 1.40625: 1.5 = 0.9375, or,

multiplying this number by 100%, we get 93.75%.

Answer: 93.75%.