Copper (II) fluoride contains 37.42% F. Given this information, determine the mass of fluoride in 70.23 g of copper (II) fluoride.
massFluoride= 70.23(1-.3742)grams
So would the answer be 43.95g?
I agree with that answer.
To determine the mass of fluoride in 70.23 g of copper (II) fluoride, we need to use the percentage composition of fluorine in the compound.
In this case, it is given that copper (II) fluoride contains 37.42% F. This means that for every 100 g of copper (II) fluoride, 37.42 g is due to the presence of fluorine.
To find the mass of fluorine in 70.23 g of copper (II) fluoride, we can set up a proportion:
(37.42 g F / 100 g CuF2) = (x g F / 70.23 g CuF2)
Cross-multiplying the equation, we get:
37.42 g F * 70.23 g CuF2 = x g F * 100 g CuF2
x g F = (37.42 g F * 70.23 g CuF2) / 100 g CuF2
Calculating this, we get:
x g F = 26.26 g F
Therefore, the mass of fluoride in 70.23 g of copper (II) fluoride is 26.26 g.