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3 years ago

Determine the molecular formula for the unknown if the molecular mass is 60.0 amu and the empirical formula is ch2o.

Chemistry

1 answer:

mezya [45]3 years ago

5 0

Answer:

The molecular formule for this unknow molecule is C2H4O2

Explanation:

The empirical formula is CH2O ( or better said CnH2nOn)

This means there are 3 elements in the formula of this molecule

⇒ Carbon (C) with a Molar mass of 12 g/mole

⇒ Hydrogen (H) with a Molar mass of 1 g/mole

⇒ Oxygen (O) with a Molar mass of 16 g/mole

We can also notice that the amount of hydrogen should 2x the amount of carbon ( also 2x the amount of oxygen).

The mass of the empirical formule = 12g/ mole + 2* 1 g/mole + 16 g/mole = 30 g/mole

To know what number is n in CnH2nOn we should divide the molecular mass by the empirical mass:

60 g/mole / 30g/mole = 2

this means n = 2

and this will give a molecular formule of C2H4O2

We can control this to calculate the molecular mass:

2*12 + 4* 1 + 2*16 = 24 + 4 + 32 = 60 g/mole

The molecular formule for this unknow molecule is C2H4O2

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2 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

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Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

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