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Kruka [31]
4 years ago
11

A sample of a compound contains 60.0 g C and 5.05 g H. Its molar mass is 78.12 g/mol. What is the compound’s molecular formula?

CH C2H2 C6H6 C6H
Chemistry
2 answers:
Inessa [10]4 years ago
5 0

A sample of a compound contains 60.0 g C and 5.05 g H.

divide by molar mass of C(12) and H(1) to get molar ratio

C: 60/12=5 and H: 5/1=5

so C:H=5:5=1:1

total molar mass=78

divide by 1C+1H to find the formula: 78/(12+1)=78/13=6

compound is C6H6


BartSMP [9]4 years ago
3 0
Step 1: Calculate Moles for each Element:

Mole of C  =  60.0 / 12  =  5

Mole of H  =  5.05 / 1.008  =  5.009

Step 2: Calculate Mole Ratio:
                                             Divide each mole value with least mole value.

                                 H     :      C

                     5.009/5       :     5/5

                                  1    :     1
So,
           Empirical Formula  =  C₁H₁ or CnHn      ------ (1)

Step 3: Calculate Molecular Formula:
By using formula,
                               n  =  Molecular Mass / Empirical Formula Mass

Molecular Mass  =  78.12 g/mol

Empirical Formula Mass  =  12 + 1  = 13          ∴ C = 12 & H = 1
So,
                               n  =  78.12 / 13 

                               n  =  6

Putting value of n in eq. 1,
  
                                                  C₆H₆
Result:
            
The correct answer is C₆H₆.
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          537.21 mol + 1074.42 mol + 2148.84 mol

        = 3760.47 mol

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                       PV = nRT

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Please Help!!
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Answer:

1. By Pressure factor: if we double the pressure volume become half of its original

2. 2.14 L

3. 2.15 L

Explanation:

part 1

Data Given:

volume of container change

temperature of remain constant

The pressure doubles

Solution:

This problem can be explained by Boyle's Law that at constant temperature pressure and volume has an inverse relation with each other.

So the volume change due to change in Pressure.

            P1V1 = P2V2

if we consider conditions at STP, as follows

initial volume V1 = 22.42 L

and

initial pressure P1 = 1 atm

if the pressure doubles then

final pressure P2 = 2 atm

Put values in Boyle's law equation

     (1 atm) (22.42L) = (2 atm) (V2)

Rearrange the above equation to find V2

           V2 =    (1 atm) (22.42L) / 2 atm

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So it is clear from calculation if we double the pressure volume become half of its original. So its the pressure due to volume become effected and decrease by its increase.

_____________

Part 2

Data Given:

Initial temperature T1= 250 K

final Temperature T2= 350 K

initial volume V1 =  ?

final volume V2 = 3.0 L

Solution:

This problem will be solved by Charles' Law equation at constant pressure

      V1 / T1 = V2 / T2 . . . . . . . . (1)

put values in above equation

      V1 / 250 K = 3.0 L / 350 K

Rearrange the above equation to calculate V1

       V1  = (3.0 L / 350 K) x 250 K

       V1  = (0.0086 L . K) x 250 K

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So the initial volume = 2.14 L

_________________

part 3

Data Given:

Initial temperature T1= 20 ºC

Convert Temperature from ºC to Kelvin

T = ºC + 273

T = 20 + 273 = 293 K

final Temperature T2= -5.00 ºC

Convert Temperature from ºC to Kelvin

T = ºC + 273

T = - 5.00 + 273 = 268 K

initial volume V1 =  2.35 L

final volume V2 = ?

Solution:

This problem will be solved by Charles' Law equation at constant pressure

      V1 / T1 = V2 / T2 . . . . . . . . (1)

put values in above equation

      2.35 L / 293 K = V2 / 268 K

Rearrange the above equation to calculate V1

       V2  = (2.35 L / 293 K) x 268 K

       V2  = (0.008 L . K) x 268 K

       V2  = 2.15 L

So the volume at -5.00ºC = 2.15 L

6 0
4 years ago
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