Sodium hydroxide was added to 120 g of a copper nitrate solution, and 4.9 g of a precipitate fell out

Sodium hydroxide was added to 120 g of a copper nitrate solution, and 4.9 g of a precipitate fell out. calculate the mass fraction of salt in the original solution.

Given:
m p (Cu (NO3) 2) = 120 g
m (Cu (OH) 2) = 4.9 g
Find: w (Cu (NO3) 2)
Decision:
Cu (NO3) 2 + 2NaOH = Cu (OH) 2 + 2NaNO3
n (Cu (OH) 2) = m / M = 4.9 g / 98 g / mol = 0.05 mol
n (Cu (NO3) 2): n (Cu (OH) 2) = 1: 1
n (Cu (NO3) 2) = 0.05 mol
m in (Cu (NO3) 2) = n * M = 0.05 mol * 188 g / mol = 9.4 g
w (Cu (NO3) 2) = m w / m p = 9.4 g / 120 g / mol * 100% = 7.83%
Answer: 7.83%