How many grams of FeSO4 are present in a 20 mL sample of a 0.500 M solution?
A) 6.08 g B) 3.04 g C) 0.760 g D) 65.8 g E) 1.52 g
What is the correct formula for this problem. I tried dividng mL by M
1 Answer
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0.5 M solution means that there is 1/2 of a mole of FeSO4 mixed into a liter of water. So, our first trick is to find the mass of 1 liter of a 0.5 M solution
Mass of Iron: 55.845 grams/mole
Mass of Sulfur: 32.065 grams/mole
Mass of Oxygen: 16 grams/mole
55.845 + 32.065 + 16 * 4 =>
151.91
So, one mole of FeSO4 has a mass of 151.91 grams. Half a mole would therefore have a mass of 75.955 grams
1 liter of water is 1000 milliliters, so we set up a ratio:
75.955 grams / 1000 ml = x grams / 20 ml
20 * 75.955 / 1000 = x
x = 75.955 / 50
x = 1.5191
E) 1.52 grams
