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tankabanditka [31]
3 years ago
7

A 14.60g sample of an unknown compound, composed only of carbon, hydrogen, and oxygen, produced 28.6g of CO2 and 14.6g of H2O in

a combustion analysis. What is the mass percent composition of oxygen of the unknown compound?
Chemistry
2 answers:
o-na [289]3 years ago
8 0

Answer: The empirical formula for the given compound is C_{2}H_{5}O

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=28.6g

Mass of H_2O=14.6g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 28.6 g of carbon dioxide, \frac{12}{44}\times 28.6=7.8g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.6 g of water, \frac{2}{18}\times 14.6=1.6 of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (14.60) - (7.8 + 1.6) = 5.2 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{7.8g}{12g/mole}=0.65moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.6g}{1g/mole}=1.6moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{5.2g}{16g/mole}=0.32moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.32 moles.

For Carbon = \frac{0.65}{0.32}=2.03\approx 2

For Hydrogen  = \frac{1.6}{0.32}=5

For Oxygen  = \frac{0.32}{0.32}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 1

Hence, the empirical formula for the given compound is C_{2}H_{5}O_{1}=C_{2}H_{5}O

leva [86]3 years ago
6 0

Answer:

\large \boxed{35.4 \, \%}

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of oxygen in 14.60 g of X.

(a) Mass of C

\text{Mass of C} = \text{28.6 g CO}_{2}\times \dfrac{\text{12.01 mg C}}{\text{44.01 mg }\text{CO}_{2}}= \text{7.805 g C}

(b) Mass of H

\text{Mass of H} = \text{14.6 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{1.633 g H}

(c) Mass of O

Mass of O = 14.60 - 7.805 - 1.633 = 5.162 g

2. Calculate the mass percent of oxygen

\text{% of oxygen} = \dfrac{\text{5.162 g}}{\text{14.60 g}} \times 100 \, \% = 35.4 \, \%\\\\\text{The compound contains $\large \boxed{\mathbf{35.4 \, \%}}$ oxygen by mass}}

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If the energy of photon emitted from the hydrogen atom is 4.09 x 10-19 J, what is
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Answer:

486 nm

Explanation:

From the question given above, the following data were obtained:

Energy (E) = 4.09×10¯¹⁹ J

Wavelength (λ) =?

Next, we shall determine the frequency of the photon. This can be obtained as follow:

Energy (E) = 4.09×10¯¹⁹ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

4.09×10¯¹⁹ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 4.09×10¯¹⁹ / 6.63×10¯³⁴

f = 6.17×10¹⁴ Hz

Next, we shall determine the wavelength of the photon. This can be obtained as follow:

Frequency (f) = 6.17×10¹⁴ Hz

Velocity of photon (v) = 3×10⁸ m/s

Wavelength (λ) =?

v = λf

3×10⁸ = λ × 6.17×10¹⁴

Divide both side by 6.17×10¹⁴

λ = 3×10⁸ / 6.17×10¹⁴

λ = 4.86×10¯⁷ m

Finally, we shall convert 4.86×10¯⁷ m to nm. This can be obtained as follow:

1 m = 1×10⁹ nm

Therefore,

4.86×10¯⁷ m = 4.86×10¯⁷ m × 1×10⁹ nm / 1 m

4.86×10¯⁷ m = 486 nm

Therefore, the wavelength of the photon is 486 nm

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