Chemistry, 09.07.2019 08:00 codycollier

Cyclohexene, a hydrocarbon, has a molar mass of 82.1 g/mole. if the combustion of 0.8300 g cyclohexene produces 0.9102 g h2o and 2.668 g co2, what is the molecular formula of this compound?

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Answer from: shanilafaridor97hl

Answer is: the molecular formula is C₆H₁₀.

Chemical reaction: C₆Hₓ + yO₂ → 6CO₂ + x/2H₂O.

m(H₂O) = 0.9102; mass of the water.

n(H₂O) = m(H₂O) ÷ m(H₂O)

n(H₂O) = 0,9102 g ÷ 18 g/mol

n(H₂O) = 0,0505 mol.

From chemical reaction: n(H) = 2 · n(H₂O).

n(H) = 0.101 mol.

m(CO₂) = 2.668 g; mass of carbon dioxide.

n(CO₂) = 2.668 g ÷ 44 g/mol.

n(CO₂) = 0.0606 mol; amount of the substance.

n(CO₂) = n(C) = 0.0606 mol.

n(C) : n(H) = 0.606 mol : 0.101 mol.

n(C) : n(H) = 6 : 10; proportion of carbon and hydrogen atoms in cyclohexene.

Answer from: calebabaltimore

The molecular formula will be C6H10.

Explanation;

Given Data:

mass of compound = 0.8300 g mass of H2O = 0.9102 g mass of CO2 = 2.668 g

Solution:

% of C = (mass of CO2/mass of compound)× (12/44)×100

% of C = (2.66÷0.8300)×(12/44)× 100 = 87.40 %

% of H = (mass of H2O/mass of compound)× (2/18)×100

% of H = (0.910÷0.8300)× (2/18)×100 = 12.18 %

Niiw we calculate moles of C and H from % as follow

moles of C = 87.40/12 = 7.2 mol

moles of H = 12.18/1 = 12.18 mol

In we take atomic ratio

C : H

12.18/7.2 : 7.2/7.2

1.69 : 1

multiply atomic ratio by 3 to get whole no.

C : H = 3 (1.69:1) = 5 : 3

So empirical formula is C3H5.

Calculations of molecular formula:

empirical formula mass of C3H5 = 41

n = molar mass of compound /empirical formula mass

n = 82.1/41 = 2

Molecular mass = n × empirical formula = 2 × C3H5 = C6H10

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

The answer is letter c

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Cyclohexene, a hydrocarbon, has a molar mass of 82.1 g/mole. if the combustion of 0.8300 g cyclohexe...