Chemistry : asked on johnsnow69
 17.06.2021

Given the empirical formula C2H3O, what is the molecular formula that has a mass of 214.8g?

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09.07.2023, solved by verified expert
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Given the empirical formula, the molecular formula that has a mass of 214.8g is C₁₀H₁₅O₅

Empirical & ,molecular formulae  

From the question, we are to determine the molecular formula

From

Molecular formula = (Empirical formula)n

Where

Given the empirical formula C2H3O, what is the, №18010633, 17.06.2021 17:38

From the given information,

Empirical formula is C₂H₃O

Therefore, Empirical formula mass = (12×2) + (1×3) + (16)

Empirical formula mass = 24 + 3 + 16

Empirical formula mass = 40 g

Then,

Given the empirical formula C2H3O, what is the, №18010633, 17.06.2021 17:38

n = 5.37

n ≅ 5

Therefore,

Molecular formula = (C₂H₃O)₅

Molecular formula = C₁₀H₁₅O₅

Hence, given the empirical formula, the molecular formula that has a mass of 214.8g is C₁₀H₁₅O₅

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Chemistry
Step-by-step answer
P Answered by Master

Given the empirical formula, the molecular formula that has a mass of 214.8g is C₁₀H₁₅O₅

Empirical & ,molecular formulae  

From the question, we are to determine the molecular formula

From

Molecular formula = (Empirical formula)n

Where

n = \frac{Molecular\ formula\ mass}{Empirical\ formula\ mass}

From the given information,

Empirical formula is C₂H₃O

Therefore, Empirical formula mass = (12×2) + (1×3) + (16)

Empirical formula mass = 24 + 3 + 16

Empirical formula mass = 40 g

Then,

n = \frac{214.8}{40}

n = 5.37

n ≅ 5

Therefore,

Molecular formula = (C₂H₃O)₅

Molecular formula = C₁₀H₁₅O₅

Hence, given the empirical formula, the molecular formula that has a mass of 214.8g is C₁₀H₁₅O₅

Learn more on Empirical and molecular formulae here:

Chemistry
Step-by-step answer
P Answered by Specialist

Empirical formula mass :-

\\ \tt\Rrightarrow C_2H_3O

\\ \tt\Rrightarrow 24+16+3=43g/mol

n:-

Molar mass/Empirical formula mass

\\ \tt\Rrightarrow \dfrac{172}{43}

\\ \tt\Rrightarrow 4

Molecular formula

4(C_2H_3O)=C_8H_12O_4
Chemistry
Step-by-step answer
P Answered by Specialist

Empirical formula mass :-

\\ \tt\Rrightarrow C_2H_3O

\\ \tt\Rrightarrow 24+16+3=43g/mol

n:-

Molar mass/Empirical formula mass

\\ \tt\Rrightarrow \dfrac{172}{43}

\\ \tt\Rrightarrow 4

Molecular formula

4(C_2H_3O)=C_8H_12O_4
Chemistry
Step-by-step answer
P Answered by PhD

17) d. C_2H_6

18. c. The empirical formula of a compound can be twice the molecular formula.

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.  

To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{molecular mass}}{\text{empirical mass}}

The empirical mass can be calculated from empirical formula and molar mass must be known.

17. Thus the empirical formula of C_2H_6 should be CH_3

18. The molecular formula will either be same as empirical formula or is a whole number multiple of empirical formula. Thus the empirical formula of a compound can never be twice the molecular formula.

Chemistry
Step-by-step answer
P Answered by PhD

17) d. C_2H_6

18. c. The empirical formula of a compound can be twice the molecular formula.

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.  

To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{molecular mass}}{\text{empirical mass}}

The empirical mass can be calculated from empirical formula and molar mass must be known.

17. Thus the empirical formula of C_2H_6 should be CH_3

18. The molecular formula will either be same as empirical formula or is a whole number multiple of empirical formula. Thus the empirical formula of a compound can never be twice the molecular formula.

Chemistry
Step-by-step answer
P Answered by Master

The molecular formula of the compound :C_4H_6O_2

Explanation:

The empirical formula of the compound =C_2H_3O

The molecular formula of the compound =C_{2n}H_{3n}O_n

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 86 g/mol

Mass of empirical formula = 43 g/mol

Putting values in above equation, we get:

n=\frac{86g/mol}{43g/mol}=2

The molecular formula of the compound :

C_{2\times 2}H_{3\times 2}O_2=C_4H_6O_2

Chemistry
Step-by-step answer
P Answered by Specialist

The molecular formula of the compound :C_4H_6O_2

Explanation:

The empirical formula of the compound =C_2H_3O

The molecular formula of the compound =C_{2n}H_{3n}O_n

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 86 g/mol

Mass of empirical formula = 43 g/mol

Putting values in above equation, we get:

n=\frac{86g/mol}{43g/mol}=2

The molecular formula of the compound :

C_{2\times 2}H_{3\times 2}O_2=C_4H_6O_2

Chemistry
Step-by-step answer
P Answered by PhD

The molecular formula is C12H18O3

Explanation:

Step 1: Data given

The empirical formula is C4H6O

Molecular weight is 212 g/mol

atomic mass of C = 12 g/mol

atomic mass of H = 1 g/mol

atomic mass of O = 16 g/mol

Step 2: Calculate the molar mass of the empirical formula

Molar mass = 4* 12 + 6*1 +16

Molar mass = 70 g/mol

Step 3: Calculate the molecular formula

We have to multiply the empirical formula by n

n = the molecular weight of the empirical formula / the molecular weight of the molecular formula

n = 70 /212 ≈ 3

We have to multiply the empirical formula by 3

3*(C4H6O- = C12H18O3

The molecular formula is C12H18O3

Chemistry
Step-by-step answer
P Answered by PhD

The molecular formula is C12H18O3

Explanation:

Step 1: Data given

The empirical formula is C4H6O

Molecular weight is 212 g/mol

atomic mass of C = 12 g/mol

atomic mass of H = 1 g/mol

atomic mass of O = 16 g/mol

Step 2: Calculate the molar mass of the empirical formula

Molar mass = 4* 12 + 6*1 +16

Molar mass = 70 g/mol

Step 3: Calculate the molecular formula

We have to multiply the empirical formula by n

n = the molecular weight of the empirical formula / the molecular weight of the molecular formula

n = 70 /212 ≈ 3

We have to multiply the empirical formula by 3

3*(C4H6O- = C12H18O3

The molecular formula is C12H18O3

Chemistry
Step-by-step answer
P Answered by Master
Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarity * L
mol = 0.15 * 0.03 = 0.0045 mol
Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarit

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