The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

Hafnium = 55.7% = 55.7g
Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

Hafnium = 0.312 ÷ 0.312 = 1
Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

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7 0

1 year ago

Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO2 and 4.96g of H2O. What is the empirical formula for the hydr

vlabodo [156]

Let empirical formula for hydrocarbon is CxHy

it will undergo combustion as

CxHy + (x + y/4) O2 ---> xCO2 + (y/2 )H2O

Given that mass of CO2 produced = 9.69 g

So moles of CO2 produced = 9.69 / 44 = 0.22 moles

So moles of carbon present = 0.22 moles

mass of H2O produced = 4.96 g

Moles of H2O produced = mass / molar mass = 4.96 / 18 = 0.28 moles

So moles of H present = 2 X 0.28 = 0.56 moles

Let us divided the moles of each with lowest value of moles

Moles of Carbon = 0.22 / 0.22 = 1 moles

moles of H = 0.56 / 0.22 = 2.55

Multiplying with two to get whole number

the ratio of carbon and hydrogen will be : C:H = 2:5

empirical formula : C2H5

4 0

3 years ago

Read 2 more answers

When methane is burned with oxygen, the products are carbon dioxide and water. if you produce 9 grams of water and 11 grams of c

Alina [70]

4 grams of methane is <span>burned with oxygen,. Hope this helped</span>

3 0

3 years ago