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3 years ago

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A crime lab received a 235-gram sample. The sample had a molecular mass of 128.1 grams and the empirical formula is CH2O. How ma

ny grams of each element are in the sample?

1 answer:

kvv77 [185]3 years ago

7 0

Explanation:

Given parameters:

Mass of sample = 235g

Molecular mass of sample = 128.1g

Empirical formula = CH₂O

Unknown:

Mass of each element in the sample = ?

Solution:

To solve this problem, we must know that the empirical formula of any compound is the simplest ratio of the atoms it contains. This is not the true formula of the compound.

Molecular formula = (Empirical formula)ₙ

Let us find the molecular mass of the sample;

CH₂O = 12 + 2(1) + 16 = 30g

128.1 = (30)n

n = 4

The molecular formula of the compound is; (CH₂O)₄ = C₄H₈O₄

Now to find the grams of each element in the sample;

Express the molecular mass of each element and that of the compound as a fraction and multiply with the given mass;

For C;

$\frac{4 x 12}{128.1} x 235\\$ = 88.06g

H; $\frac{8 x 1}{128.1} x 235$ = 14.68g

O: $\frac{4 x 16}{128.1} x 235$ = 117.41g

learn more:

Mass composition brainly.com/question/3018544

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At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

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Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

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