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3 years ago

A compound consists of 81.8% C and 18.2% H by mass. What is its empirical formula?

Chemistry

1 answer:

nataly862011 [7]3 years ago

5 0

Answer:

Empirical formula of 81.8 carbon and 18.2 hydrogen

Steps

Find the empirical formula of a compound that has 81.8% of Carbon (C), 18.2% of Hydrogen (H).
Divide by fractional component of each mole value: C=1.5354936431103, H=4.0709872862205Round to closest whole numbers: C=2, H=4

<h3> Combine to get the empirical formula: C2H4</h3>

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3 years ago

A sample in the laboratory is found to contain 3.36 grams of hydrogen, 20.00 grams of carbon, and 26.64 grams of oxygen. The mol

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Answer:

Empirical formula is CH₂O.

Molecular formula = C₆H₁₂O₆

Explanation:

Given data:

Mass of hydrogen = 3.36 g

Mass of carbon = 20.00 g

Mass of oxygen = 26.64 g

Molar mass of compound = 180.156 g/mol

Empirical formula = ?

Molecular formula = ?

Solution:

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number

Number of gram atoms of H = 3.36 / 1.01 = 3.3

Number of gram atoms of O = 26.64 / 16 = 1.7

Number of gram atoms of C = 20 / 12 = 1.7

Atomic ratio:

C : H : O

1.7/1.7 : 3.3/1.7 : 1.7/1.7

1 : 2 : 1

C : H : O = 1 : 2 : 1

Empirical formula is CH₂O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = CH₂O = 12×1 + 2× + 16

Empirical formula mass = 30

n = 180.156 / 30

n = 6

Molecular formula = n (empirical formula)

Molecular formula = 6 (CH₂O)

Molecular formula = C₆H₁₂O₆

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3 years ago

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Explanation:

number of moles

= number of atoms ÷ Avogadro's constant

Avogadro's constant= 6.022 ×10²³

Thus, number of moles

= 1.25×10²⁴ ÷ (6.022 ×10²³)

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3 years ago

What is the molarity (M) of chloride ions in a solution prepared by mixing 155 ml of 0.276 M calcium chloride with 384 ml of 0.4

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Answer: The concentration of $Cl^-$ ions in the resulting solution is 1.16 M.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the $CaCl_2$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $AlCl_3$

We are given:

$n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL$

Putting all the values in above equation, we get

$M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M$

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Hence, the concentration of $Cl^-$ ions in the resulting solution is 1.16 M.

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3 years ago

A compound consists of 81.8% C and 18.2% H by mass. What is its empirical formula?​

A compound consists of 81.8% C and 18.2% H by mass. What is its empirical formula?