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jolli1 [7]
1 year ago
12

A compound has a percent composition of 81.71% C and 18.29% H. What is the empirical formula of this compound?

Chemistry
1 answer:
SCORPION-xisa [38]1 year ago
4 0

Considering the definition of empirical formula, the empirical formula is C₃H₈.

<h3>Definition of empirical formula</h3>

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

<h3>Empirical formula in this case</h3>

In this case, in first place you know the percent composition:

  • C: 81.71 %
  • H: 18.29%

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have 81.71 grams of carbon and 18.29 grams of hydrogen H.

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

  • C: \frac{81.71 grams}{12\frac{grams}{mole} }= 6.81 moles
  • H:\frac{18.29 grams}{1\frac{grams}{mole} }= 18.29 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

  • C: \frac{6.81 moles}{6.81 moles}= 1
  • H:\frac{18.29 moles}{6.81 moles}= 2.68 ≅ \frac{8}{3}

To express this relationship in the form of simple integers, it is necessary to multiply by a simple number to achieve this:

  • C: 1×3  =3
  • H:≅ \frac{8}{3}×3= 8

Therefore the C: H mole ratio is 3: 8

Finally, the empirical formula is C₃H₈.

Learn more about empirical formula:

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#SPJ1

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Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ (n_{A}) = 1

Mole ratio of the base, NaOH (n_{B}) = 2

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH (M_{B}) = 0.505 M

Volume of the acid, H₂SO₄ (V_{A}) = 40 mL

Molarity of the acid, H₂SO₄ (M_{A}) = 0.505 M

<h3>Volume of the base, NaOH (V_{B}) =? </h3>

\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}

Cross multiply

0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4

Divide both side by 0.505

V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}

Therefore, the volume of NaOH required is 0.08 dm³

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Answer:

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