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10 months ago

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

Chemistry

1 answer:

uranmaximum [27]10 months 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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2 years ago

Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.

Vsevolod [243]

The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

The mole ratio of the acid, H₃PO₄ (nA) = 2
The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>

Molarity of acid, H₃PO₄ (Ma) = 0.390 M
Volume of acid, H₃PO₄ (Va) = 24.5 mL
Molarity of base, Ca(OH)₂ (Mb) = 0.279 M

Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

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Cross multiply

2 × 0.279 × Vb = 9.555 × 3

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Divide both side by 0.558

Vb = 28.665 / 0.558

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Learn more about titration:

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