A compound is found to contain 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen by weight. What is the empirical formula for this compound

Empirical formula is C₄H₁₀O

Explanation:

Values for C, H and O are determined as centesimal composition.

64.80 g of C in 100g of compound

13.62g of H in 100 g of compound

21.58 g of O in 100 g of compound.

We convert the mass to moles:

64.80 g . 1mol/ 12g = 5.4 moles of C

13.62 g . 1 mol /1g = 13.62 moles of H

21.58 g . 1 mol/16g = 1.35 moles of O

We pick the lowest value and we divide:

5.4 moles of C / 1.35 = 4 C

13.62 moles of H / 1.35 = 10 H

1.35 moles of O / 1.35 = 1 O

Empirical formula is C₄H₁₀O, it can be the diethyl ether.

We confirm, the excersise is well done.

Molar mass = 74g/mol

74 g of compound we have (12 . 4)g of C

In 100 g of compound we may have (100 . 48) / 74 = 64.8 g

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

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