What is the percentage of water to the nearest tenth in this compound Na2CO3 x 10H2O?

Answer 1

there are 159.6g of cupric sulfate and 90.1g of waters63.5gCu+32.1gS+(16.0*4)gO=159.6gCuSO4 5(1.01*2)gH+5(16.0)gO=90.1g5H20

therefore, the gram formula mass is 249.7g

the percent composition of water is 36.1%((90.1g)/(249.7g))*100%

p.s.-my basic math skills suck so you may want to take the given values and crank them through the equation to check

Your math is correct. I ran the problem myself and got the same answer. We even did this as a lab in Chemistry class and I got near the same answer (35% in an unregulated test).

Answer 2

The compound Na2CO3 x 10H2O has a total of 10 water molecules. The percentage of water in this compound is calculated by dividing the molar mass of water (18 g/mol) by the total molar mass of the compound, then multiplying by 100. Therefore, the percentage of water in Na2CO3 x 10H2O is 46.7%.

Answer 3

CuSO4 x5H20 is the molecular formula for cupric sulfate pentahydrate. Its molecular weight of 247.9 grams per mole. Also, the molecular weight of water is 18.02 grams per mole. The molecule is a pentahydrate, meaning it has 6 waters. This means 108.1 grams of each cupric sulfate pentahydrate is composed of water. 90.1 grams of water divided by the total mass of 249.7 equals .36. The answer then is 36% water.

Answer 4

The molar mass of CuSO4 is 159.61 grams per mole. The molar mass of 5 moles of water is 90.1 grams. Then water makes up 36.1 percent of the crystal.

Answer 5

Mass of h2o- 5X18 = 90 mass of cuso4=160g thus, % of h2o= 90/160x100=56.2%

Answer 6

The molar mass of Na2CO3 is 105.988 g/mol. The molar mass of H2O is 18.0153 g/mol. The total weight of the compound then is 286.141 grams. Water then comprises 62.96 percent of the compound.

Answer 7

The percentage of water is 62,9 %.

Answer 8

36.08