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tekilochka [14]
3 years ago
14

Benzene contains only carbon and hydrogen and has a molar mass of 78.1 g/mol. Analysis shows the compound to be 7.44% hydrogen b

y mass. Find the empirical and molecular Formula of benzene.
Chemistry
1 answer:
Sindrei [870]3 years ago
6 0

Hydrogen - 7.44%

Carbon - (100-7.44)% = 92.56%


Lets take 100 g of benzene, then we have

Hydrogen - 7.44 g

Carbon - 92.56 g


n - number of moles

n(H) = 7.44g *1 mol/1.0g = 7.44 mol

n(C) = 92.56 g* 1mol/12.0 g ≈ 7.713 mol


n(C) : n(H) = 7.713 mol : 7.44 mol = 1:1

Empirical formula is CH.


M(CH) = (12.0+ 1.00) g/mol = 13.0 g/mol

M (benzene) = 78.1 g/mol

M (benzene)/M(CH)= 78.1 g/mol/13.0 g/mol = 6

So, molecular formula of benzene is C6H6.


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A certain substance X has a normal freezing point of −3.1°C and a molal freezing point depression constant =Kf·6.23°C·kgmol−1. C
muminat

Answer:

Freezing T° of solution =  - 7.35 °C

Explanation:

This is about the freezing point depression, a colligative property which depends on solute.

The formula is: Freezing T° pure solvent - Freezing T° solution = m . Kf . i

Freezing T° of pure solvent is -3.1°C

At this case, i = 1. As an organic compound the urea does not ionize.

We determine the molality (mol/kg of solvent)

We convert mass to moles:

12.3 g . 1mol / 60.06 g = 0.205 moles

0.205 mol / 0.3 kg = 0.682 mol/kg

We replace data in the formula:

-3.1°C - Freezing T° of solution  = 0.682 mol/kg . 6.23 kg°C/mol . 1

Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1 + 3.1°C

Freezing T° of solution =  - 7.35 °C

6 0
4 years ago
A 8.249 gram sample of copper is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 13.18 g. D
levacccp [35]

Answer:

CuF_2 the empirical formula of the metal fluoride.

Explanation:

Mass of copper heated = 8.249 g

Mass of copper fluoride formed = 13.18 g

Mass of fluorine gas in copper fluoride = x

13.18 g = 8.249 g + x\\x= 13.18 - 8.249 g = 4.931 g

Moles of copper :

= \frac{8.249 g}{63.546 g/mol}=0.1298 mol

Moles of fluorine:

= \frac{4.931 g}{18.998 g/mol}=0.2596 mol

For the empirical formula divide the smallest mole of an element with all the moles of elements present in the compound.

Copper= \frac{0.1298 mol}{0.1298 mol}=1\\Fluorine = \frac{0.2596 mol}{0.1298 mol}=2

The empirical formula of the copper fluoride = CuF_2

CuF_2 the empirical formula of the metal fluoride.

6 0
3 years ago
Complete the sentence. A substance that is more basic has a___pH
alex41 [277]

Answer:

Higher

Explanation:

6 0
3 years ago
A substance that makes things taist sweeter
elena-s [515]
Sugar. (We need a design tech section)
5 0
4 years ago
KBr will undergo dissociation in solution while benzene will undergo solvation.
Degger [83]
Kbr will undergo dissociation in an aqueous solution. But Benzene will not dissolve in water so it will be a solvent.

hoping this help
8 0
3 years ago
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