Chemistry, 09.03.2020 16:32 matt16913

The percent composition by mass of an unknown compound with a molecular mass of 60.052 amu is 40.002% C, 6.7135% H, and 53.284% O. Determine the compound's empirical and molecular formulas. Empirical formula:

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The empirical formula and molecular formula of the compound with a molecular mass of 60.052 amu containing 40.002% C, 6.7135% H, and 53.284% O are:

1. The empirical formula of the compound is CH₂O

2. The molecular formula of the compound is C₂H₄O₂

1. Determination of the empirical formula of the compound.

C = 40.002%

H = 6.7135%

O = 53.284%

Divide by their molar mass

C = 40.002 / 12 = 3.3335

H = 6.7135 / 1 = 6.7135

O = 53.284 / 16 = 3.33025

Divide by the smallest

C = 3.3335 / 3.33025 = 1

H = 6.7135 / 3.33025 = 2

O = 3.33025 / 3.33025 = 1

Thus, the empirical formula of the compound is CH₂O

2. Determination of the molecular formula of the compound

Empirical formula = CH₂O

Molar mass = 60.052 amu

Molecular formula =?

Molecular formula = Empirical × n = molar mass

[CH₂O]n = 60.052

[12 + (2×1) + 16]n = 60.052

30n = 60.052

Divide both side by 30

n = 60.052 / 30

n = 2

Molecular formula = [CH₂O]n

Molecular formula = [CH₂O]₂

Molecular formula = C₂H₄O₂

Therefore, the molecular formula of the compound is C₂H₄O₂

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Answer from: misswonderless

Empirical formula and molecular formula are CH_2O and C_2H_4O_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 40.002 g

Mass of H= 6.7135 g

Mass of O = 53.284 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{ 40.002g}{12g/mole}=3.3335moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{ 6.7135g}{1g/mole}=6.7135moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.284 g}{16g/mole}=3.3302moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.3335}{3.3302}=1$

For H = $\frac{6.7135}{3.3302}=2$

For O = $\frac{3.3302}{3.3302}=1$

The ratio of C : H : O= 1 : 2 : 1

Hence the empirical formula is CH_2O

The empirical weight of CH_2O = 1(12)+2(1)+1(16)= 30 g.

The molecular weight = 60.052 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{ 60.052 }{30}=2$

The molecular formula will be= $2\times CH_2O=C_2H_4O_2$

Thus empirical formula and molecular formula are CH_2O and C_2H_4O_2

Answer from: Quest

pure substance: iron,copper

impure substance: water,charcoal

Answer from: Quest

answer: any goods being imported from another country may be subject to issued by the federal government are tariffs

explanation:

tariffs are a tax on anything that are imported from abroad. government use them to protect and boost their local industries and the thought is that they encourage consumers to buy local instead of importing and they will hope will protect jobs and economy in the country that implements them

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The percent composition by mass of an unknown compound with a molecular mass of 60.052 amu is 40.002...

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