If the empirical formula is $C_{2} H_{5}$ $C_2H_5$ , and the molecular mass is $58 \cdot g \cdot m o l^{- 1}$ $58gmol^-1$ , what is the molecular formula?

Question

If the empirical formula is $C_{2} H_{5}$ $C_2H_5$ , and the molecular mass is $58 \cdot g \cdot m o l^{- 1}$ $58gmol^-1$ , what is the molecular formula?

2 Answers

Impact of this question

15593 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-28T18:16:21

If I am reading your question correctly, it is $H_{3} C - C H_{2} C H_{2} C H_{3}$ $H_3C-CH_2CH_2CH_3$ or $C_{4} H_{10}$ $C_4H_10$ .

Explanation:

Butane is a $C_{4}$ $C_4$ alkane. The isomer depicted is the straight chain, n -butane. It has 1 other isomer.

Answer 2 · 2015-09-29T02:00:31

$C_{4} H_{10}$ $"C"_4"H"_10$

Explanation:

I assume that you have to determine the molecular formula of butane starting from its empirical formula and its molar mass, which is $≅ 58 g$ $~= "58 g"$ .

If you started from butane's percent composition of carbon and hydrogen, and provided that your calculations were correct, then the empirical formula should have come out as $C_{2} H_{5}$ $"C"_2"H"_5$ .

Now, you know that the empirical formula tells you the simplest ratio of atoms between the elements that make up a compound.

In your case, you know that a butane molecule contains 5 atoms of hydrogen for every 2 atoms of carbon.

Since you know what the compound's molar mass is, you can say that

$(2 \times M_{M carbon} + 5 \times M_{M hydrogen}) \cdot n = 58 g$ $(2 xx M_"M carbon" + 5 xx M_"M hydrogen") * color(blue)(n) = "58 g"$

You know that the molecule contains at least two atoms of carbon and five atoms of hydrogen, but you don't exactly how many of each you have $\to$ $->$ this is what $n$ $color(blue)(n)$ represents.

So, use the molar masses of carbon and hydrogen to get the value of $n$ $color(blue)(n)$

$(2 \times 12.01 g/mol + 5 \times 1.01 g/mol) \cdot n = 58 g$ $(2 xx "12.01 g/mol" + 5 xx "1.01 g/mol") * color(blue)(n) = "58 g"$

$color(blue)(n) = (58color(red)(cancel(color(black)("g"))))/(29.07color(red)(cancel(color(black)("g")))) = "1.995 ~= 2$

This means tha tthe molecular formula of butane is

$("C"_2"H"_5)_2 = color(green)("C"_4"H"_10)$

Science

Math

Humanities

... and beyond

If the empirical formula is $C_{2} H_{5}$ $C_2H_5$ , and the molecular mass is $58 \cdot g \cdot m o l^{- 1}$ $58gmol^-1$ , what is the molecular formula?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If the empirical formula is C_2H_5C2H5C_2H_5, and the molecular mass is 58*g*mol^-158⋅g⋅mol−158*g*mol^-1, what is the molecular formula?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

If the empirical formula is $C_{2} H_{5}$ $C_2H_5$ , and the molecular mass is $58 \cdot g \cdot m o l^{- 1}$ $58gmol^-1$ , what is the molecular formula?