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2 years ago

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]2 years 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.

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Oxygen and hydrogen are kept under the same temperature and pressure. Oxygen molecule is 16 times more massive than hydrogen one

lozanna [386]

Explanation:

When we increase the temperature of a substance then there will occur an increase in the kinetic energy of its molecules.

Also, K.E = $\frac{3}{2}kT$

So, kinetic energy is directly proportional to the temperature.

Hence, when temperature and pressure are kept the same for both oxygen and hydrogen gas then values of their kinetic energy will be the same irrespective of their masses.

Thus, we can conclude that kinetic energy of oxygen molecule is the same as compared to hydrogen molecule.

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3 years ago

Calculate the number of grams of CO that can react with 0.400 kg of Fe2O3.

aleksandr82 [10.1K]

Answer:

mass of CO = 210.42 g

mass in three significant figures = 210. g

Explanation:

Given data:

mass of Fe2O3 = 0.400 Kg

mass of CO= ?

Solution:

chemical equation:

Fe2O3 + 3CO → 2Fe + 3CO2

Now we will calculate the molar mass of Fe2O3 and CO.

Molar mass of Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol

Molar mass of CO = 12+ 16 = 28 g/mol

now we will convert the kg of Fe2O3 in g.

mass of Fe2O3 = 0.400 kg × 1000 = 400 g

number of moles of Fe2O3 = 400 g/ 159.69 g/mol = 2.505 mol

mass of CO = moles of Fe2O3 × 3( molar mass of CO)

mass of CO = 2.505 mol × 84 g/mol

mass of CO = 210.42 g

mass in three significant figures = 210. g

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3 years ago

Read 2 more answers

What is the net ionic equation for the reaction that occurs when aqueous solutions of koh and zncl2 are mixed?

Mademuasel [1]

Answer:

Net ionic equation:

Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)

Explanation:

Chemical equation:

ZnCl₂ + KOH → KCl + Zn(OH)₂

Balanced chemical equation:

ZnCl₂ + 2KOH → 2KCl +Zn(OH)₂

Ionic equation;

Zn²⁺(aq) + 2Cl⁻(aq) + 2K⁺(aq) + 2OH⁻(aq) → 2K⁺(aq) + 2Cl⁻(aq) +Zn(OH)₂(s)

Net ionic equation:

Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)

The K⁺ and Cl⁻ are spectator ions that's why these are not written in net ionic equation. The Zn(OH)₂ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

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3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

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Answer:

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