Compare the mass of table salt from different deposits if 8 grams of it contain 2.5% impurities

Compare the mass of table salt from different deposits if 8 grams of it contain 2.5% impurities and 12 grams is 1 point nine.

Given:
m1 tech. (NaCl) = 8 g
ω1 approx. = 2.5%
m2 tech. (NaCl) = 12 g
ω2 approx. = 1.9%

To find:
m1 clean (NaCl)? m2 clean (NaCl)

Decision:
1) ω1 (NaCl) = 100% – ω1 approx. = 100% – 2.5% = 97.5%;
2) m1 clean (NaCl) = ω1 (NaCl) * m1 tech. (NaCl) / 100% = 97.5% * 8/100% = 7.8 g;
3) ω2 (NaCl) = 100% – ω2 approx. = 100% – 1.9% = 98.1%;
4) m2 clean (NaCl) = ω2 (NaCl) * m2 tech. (NaCl) / 100% = 98.1% * 12/100% = 11.8 g;
5) m1 clean (NaCl) <m2 pure. (NaCl).

Answer: The second sample contains more NaCl than the first.