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Anvisha [2.4K]
3 years ago
5

Which pair shares the same empirical formula?

Chemistry
2 answers:
ZanzabumX [31]3 years ago
4 0

Answer:- 3. CH_3 and C_2H_6

Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.

For example, the molecular formula of benzene is C_6H_6 . The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.

In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and  the empirical formula is CH_2 . In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.

In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is CH_2 . The C to H ratio for second molecule is 1:4, so the empirical formula is CH_4 . Here also, the empirical formulas are not same and hence it is also not the right choice.

In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is CH_3 . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also CH_3 . Hence. this is the correct choice.

In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is CH_2 . As the empirical formulas are different, it is not the right choice.

So, the only and only correct pair is the third one. 3. CH_3 and C_2H_6

ANTONII [103]3 years ago
4 0
CH3 and C2H6 same empirical
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The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent com
FrozenT [24]
  • The mass percent of Pentane in solution is 16.49%
  • The mass percent of Hexane in solution is 83.51%

<u>Explanation</u>:

  • Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.
  • Convert these values to mol% using their molecular weights:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

  • From Raoult's law we know:  

Pp = xp \times Pp - vap = yp \times Pt                                       (1)

Ph = xh \times Ph - vap = yh \times Pt                                       (2)

  • Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) \times Ph - vap = (1 - yp) \times Pt                                   (3)

  • Substituting (1) into (3) we get:

(1-xp) \times Ph - vap = (1 - yp) \times xp \times Pp - vap / yp            (4)

  • Rearrange for xp:

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap)      (5)

  • Subbing in the values we find:

Pentane mol% in solution: xp = 19.08%  

Hexane mol% in solution: xh = 80.92%

  • Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol \times 72.15 g/mol

     = 13.766 g

mh = 0.8092 mol \times 86.18 g/mol

     = 69.737 g

  • Mass% of Pentane solution = 13.766/(13.766+69.737)

                                                       = 16.49%

  • Mass% of Hexane solution  = 83.51%
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KatRina [158]

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Explanation:

The following data were obtained from the question:

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V1 (initial volume) = 400mL

P1 (initial pressure) = 760mmHg

P2 (final pressure) = 380mmHg

V2 (final volume) = 551mL

T2 (final temperature) =?

Applying the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas can be obtained as follow:

P1V1/T1 = P2V2/T2

760 x 400/273 = 380 x 551/T2

Cross multiply to express in linear form as shown below:

760 x 400 x T2 = 273 x 380 x 551

Divide both side by 760 x 400

T2 = (273 x 380 x 551)/(760 x 400)

T2 = 188.03K.

Therefore, at a temperature of 188.03K, the gas sample will have a volume of 551mL

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