1) A compound is found to contain 34.60 % chlorine , 46.85 % oxygen , and 18.55
ID: 950041 • Letter: 1
Question
1)
A compound is found to contain 34.60 % chlorine , 46.85 %oxygen , and 18.55 % fluorine by mass.
To answer the question, enter the elements in the order presented above.
QUESTION 1:
The empirical formula for this compound is .
QUESTION 2:
The molar mass for this compound is 102.5 g/mol.
The molecular formula for this compound is . 2)
A 29.30 gram sample of copper is heated in the presence of excessfluorine. A metal fluoride is formed with a mass of 46.82 g. Determine the empirical formula of the metal fluoride.
Enter the elements in the order Cu , F
empirical formula =
4)
A 24.65 gram sample of copper is heated in the presence of excessfluorine. A metal fluoride is formed with a mass of 32.02 g. Determine the empirical formula of the metal fluoride.
Enter the elements in the order Cu , F
empirical formula =
)
are present in 3.12 grams of carbon monoxide ?
grams carbon.
2. How many GRAMS of carbon monoxide can be produced from 1.29 grams of oxygen ?
grams carbon monoxide. 5) 5)
6) forms a hydrate with four water molecules per formula unit. The empirical formula for this compound is .
Explanation / Answer
1)
A compound is found to contain 34.60 % chlorine , 46.85 %oxygen , and 18.55 % fluorine by mass.
MW Cl = 35
MW O = 16
MW F = 19
then
assume a 100 g basis:
34.60 g are Cl --> mol = massMW = 34.6/35 = 0.98857 = 1
46.85 g are O --> mol = massMW = 46.48/16= 2.905= 3
18.55 g are F --> mol = massMW = 18.55/19=0.97631= 1
empirical formula is ClO3F
2)
if
MW = The molar mass for this compound is 102.5 g/mol.
the molecular formula
MW of empirical formula = ClO3F = 100
then
MW/Mwempirical = 102.5/100 = 1.025
approx 1X; that is, the molecular formula = empiorical formula in this case
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