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

The correct name for n3-ion should be azide.

Answer from: Quest

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

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