A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?

Chemistry

1 answer:

uranmaximum [27]1 year ago

7 0

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

<h3>How to calculate empirical formula?</h3>

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.

Learn more about empirical formula at: brainly.com/question/14044066

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How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium

Sedbober [7]

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4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

HCl: 2 moles
Na: 1 mole
NaCl: 2 moles
H₂: 1 mole

You know the following masses of each element:

H: 1 g/mole
Cl: 35.45 g/mole
Na: 23 g/mole

So, the molar mass of each compound participating in the reaction is:

HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
Na: 23 g/mole
NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
H₂: 2* 1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

HCl: 2 moles* 36.45 g/mole= 72.9 g
Na: 1 mole* 23 g/mole= 23 g
NaCl: 2 moles* 58.45 g/mole= 116.9 g
H₂: 1 mole* 2 g/mole= 2 g

So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

$mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}$