Find the formula of the compound with the composition: Na – 57.5%; O – 40%; H – 2.5%.

1. Let’s find the number of elements in 100 g of the desired substance:

m (Na) = m (in-va) * ω (Na) / 100% = 100 g * / 100% = 100 g * 57.5% / 100% = 57.5 g.

n (Na) = m (Na) / M (Na) = 57.5 g / 23 g / mol = 2.5 mol.

m (O) = m (in-va) * ω (O) / 100% = 100 g * / 100% = 100 g * 40% / 100% = 40 g.

n (O) = m (O) / M (O) = 40 g / 16 g / mol = 2.5 mol.

m (H) = m (in-va) * ω (H) / 100% = 100 g * / 100% = 100 g * 2.5% / 100% = 2.5 g.

n (H) = m (H) / M (H) = 2.5 g / 1 g / mol = 2.5 mol.

2. Let’s compose the ratio of elements in the substance:

n (Na): n (O): n (H) = 2.5 mol: 2.5 mol: 2.5 mol = 1: 1: 1, hence the formula is NaOH.

Answer: NaOH