A 0.400g sample of a whit powder contains 0.414g of potassium, 0.115g of surfer, and 0.144g of oxygen. What is the empirical formula for this compound? Explain how you got your answer.
-
I assume that you mean 0.115 g of sulfur (not surfer) and that there are 0.141 g K (NOT 0.414 g). There must be 0.141 g K since the total mass of g K + g S + g O = 0.400 g. Convert the given quantities to moles by dividing g by atomic mass.
0.141 g K x (1 mole K / 39.1 g K) = 0.00361 moles K
0.115 g S x (1 mole S / 32.1 g S) = 0.00358 moles S
0.144 g O x (1 mole O / 16.0 g O) = 0.00900 moles O
Divide these three numbers by the smallest (0.00358)
K: 0.00361 / 0.00358 = 1.01
S: 0.00358 / 0.00358 = 1.00
O: 0.00900 / 0.00358 = 2.51
If we double these values to get whole numbers:
K = 2.01
S = 2.00
O = 5.02
The mole ratio between K / S / O is 2 / 2 / 5, so the empirical formula is K2S2O5.
I have never heard of this compound so I am wondering if your original data is valid.
0.141 g K x (1 mole K / 39.1 g K) = 0.00361 moles K
0.115 g S x (1 mole S / 32.1 g S) = 0.00358 moles S
0.144 g O x (1 mole O / 16.0 g O) = 0.00900 moles O
Divide these three numbers by the smallest (0.00358)
K: 0.00361 / 0.00358 = 1.01
S: 0.00358 / 0.00358 = 1.00
O: 0.00900 / 0.00358 = 2.51
If we double these values to get whole numbers:
K = 2.01
S = 2.00
O = 5.02
The mole ratio between K / S / O is 2 / 2 / 5, so the empirical formula is K2S2O5.
I have never heard of this compound so I am wondering if your original data is valid.