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Gnom [1K]
2 years ago
9

A mixture of NaCl and CuCO3 was dissolved in water and the resulting slurry was filtered in to a beaker. The beaker and the filt

er paper were dried and weighed.
Mass of container: 54.558g
Mass of container and mixture: 56.779g
Mass of filter paper: 1.176g
Mass of filter paper and residue: 2.656g
Mass of beaker: 77.575g
Mass of beaker and dried residue: 78.875g

Mass of mixture =
Mass of NaCl =
%NaCl (w/w) =
Mass of CuCO3 =
% (w/w) CuCO3 =

Do the percentages add up to 100% in this example? Why or why not? What do you think happened?
Chemistry
1 answer:
Firdavs [7]2 years ago
6 0

The percentage adds up to 100%.

<h3>Percentages</h3>

Mass of mixture = mass of container+mixture - mass of container = 56.779 - 54.558 = 2.221 g

CuCO3 is insoluble in water. Thus:

Mass of CuCO3 = mass of beaker and residue - mass of beaker = 78.875 - 77.575 = 1.300 g

Mass of NaCl = mass of mixture - mass of CuCO3 = 2.221 - 1.300 = 0.921 g

%NaCl (w/w) = weight of NaCl/weight of mixture = 10.921/2.221 = 41.468%

% (w/w) CuCO3 = weight of CuCO3/weight of mixture = 1.300/2.221 = 58.532%

Addition of percentages = 41.468 + 58.532 = 100%

More on percentages can be found here: brainly.com/question/24159063

#SPJ1

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A 2.500g sample of compound containing only Carbon and Hydrogen is found containing 2.002g of Carbon. Determine the empirical fo
Dima020 [189]

The empirical formula : CH₃

<h3>Further explanation</h3>

Given

2.5 g sample

2.002 g Carbon

Required

The empirical formula

Solution

Mass of Hydrogen :

= 2.5 - 2.002

= 0.498

Mol ratio C : H :

C : 2.002/12 = 0.167

H : 0.498/1 = 0.498

Divide by 0.167 :

C : H = 1 : 3

7 0
2 years ago
A solution contains 50.0g of heptane (C7H16)and 50.0g of octane (C8H18) at 25 degrees C.The vapor pressures of pure heptane and
AleksandrR [38]

Answer:

a)Pheptane = 24.3 torr          

Poctane = 5.12 torr    

b)Ptotal vapor = 29.42 torr

c)  81 % heptane

    19 % octane

d) See explanation below

Explanation:

The partial pressure is given by Raoult´s law as:

Pa = Xa Pºa where Pa = partial pressure of component A

                               Xa = mole fraction of A

                               Pºa = vapor pressure of pure A

For a binary solution what we have to do is compute the partial  vapor pressure of each component and then add them together to get total vapor pressure.

In order to calculate the composition of the vapor  in part b), we will first calculate the mole fraction of each component in the vapor which is given by the relationship:

          Xa = Pa/Pt where Xa = mol fraction of  in the vapor

                                       Pa = partial pressure of A as calculated above

                                        Pt = total vapor pressure

Once we have mole fractions we can calculate the masses of the components for part c)    

a)                  

 MW heptane = 100.21 g/mol

 MW octane = 114.23 g/mol

mol heptane = 50.0 g / 100.21 g/mol = 0.50 mol

mol octane = 50.0 g/ 114.23 g/mol = 0.44 mol

mol total = 0.94 ⇒ Xa= 0.50/0.94 = 0.53 and

                             Xb= 0.44/0.94 = 0.47

Pheptane = 0.53 x 45.8 torr = 24.3 torr

Poctane = 0.47 x 10.9 torr = 5.12 torr

b) Ptotal = 24.3 torr +5.12 torr = 29.42 torr

c) We will call Y the mole fraction in the vapor to differentiate it from the mole fraction in solution

Y heptane (in the vapor) = 24.3 torr/ 29.42 torr = 0.83

Y octane (in the vapor) = 5.12 torr/ 29.42 torr = 0.17

d) To solve this part   we will assume that since the molecular weights are similar then having a mole fraction for heptane of 0.82, we could say that for every mole of mixture we have 0.82 mol heptane and 0.17 mol octane  and then we can calculate the masses:

0.82 mol x 100.21  g/mol = 82.2 g

0.17 mol x 114.23 g/mol =  19.4 g

total mass = 101.6

% heptane = 82.2 g/101.6g x 100 = 81 %

% octane = 19 %

There is another way to do this more exactly by calculating the average molecular weight of the mixture:

average MW = 0.83 (100.21 g/mol)  + 0.17 ( 114.23 g/mol ) = 102. 6 g/mol

and then  having a mol fraction of 0.83  means in 1 mol of mixture we have 0.83 mol heptane and 0.17 mol octane then the masses are:

mass heptane = 0.83 x 100.21 g/mol = 83.2 g

mass octane = 0.17 x  114.23 g/mol = 19.4 g

mass of mixture = 1 mol x MW mixture = 1 mol x 102.6 g/mol 102.6 g

% heptane = (83.2 g/ 102.6 g ) x 100 g = 81 %

% octane = 100 - 81 = 19 %

d)The composition of the vapor is different from the composition of the solution because the vapor is going to be richer in the more volatile compound in the solution which in this case is heptane ( 45.8  vs 10.9 torr).

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