All categories

3 years ago

What is the molecular formula for the compound with a formula mass of 58.12 and an empirical formula of c2h5?

Chemistry

2 answers:

ira [324]3 years ago

8 0

Answer:

C₄H₁₀

Explanation:

The empirical formula is the reduced formula of the molecule. So, calling n the number of empirical formulas, the molecule formula will be:

n(C₂H₅)

The molar mass of C is 12 g/mol, and of H is 1 g/mol. So, the total mass will be:

nx(2x12 + 5x1) = 58.12

29n = 58.12

n = 2

Then, the molecular formula is:

2(C₂H₅) = C₄H₁₀

LenKa [72]3 years ago

3 0

The molecular formula is C4H10.

You might be interested in

Polygamy is which of the following? The practice of having two or more husbands. The practice of having two or more wives. The p

yuradex [85]

The practice of having two or more wives

8 0

3 years ago

Read 2 more answers

5. What is the molar mass of ethanol (C2H5OH)?

AVprozaik [17]

The molar mass is given by the sum of the atomic masses of the component elements of the substance.

In our case, the substance is ethanol or C₂H₅OH.

$A_C=12.011$

$A_H=1.0079$

$A_O=15.9994$

$\mathcal M_{C_2H_5OH} = 2 \times A_C + 6 \times A_H + A_O = 2 \times 12.011 + 6 \times 1.00079+ 15.999 4= 24.022 + 6.0474 + 15.9994= 46.069≈ \boxed{46.07 \dfrac{g}{mol}}$

5 0

3 years ago

Solve the following: Which one is more metallic

Degger [83]

Answer:

aluminum

Explanation:

7 0

4 years ago

Who is the father of genetics

-BARSIC- [3]

Gregor Mendel

Hope this helped !

6 0

3 years ago

Read 2 more answers

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0

4 years ago