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3 years ago

What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?

Chemistry

2 answers:

PtichkaEL [24]3 years ago

4 0

Answer:

The right choice is CHCl₃

Explanation:

To find the empirical formula firstly, change the mass to moles.

For Carbon :

no. of moles = (mass / molar mass) = (5.03 g / 12 g/mol )= 0.42 mol

For Hydrogen :

no. of moles = (mass / molar mass) = (0.42 g / 1 g/mol )= 0.42 mol

For Chlorine :

no. of moles = (mass / molar mass) = (44.5 g / 35.5 g/mol )= 1.25 mol

The ratio for Carbon and Hydrogen is 1 : 1

0.42 mol / 042 mol =1.00

Then find a ratio between the moles.

The ratio of Chlorine to both Carbon and Hydrogen is 3:1

1.25 mol / 0.42 mol = 2.98 ≅ 3

So the ratio is 1 C : 1 H : 3 Cl.

So, the right choice is CHCl₃

Carbon usually forms four bonds one to Hydrogen and 3 to Chlorine atoms.

Chlorine usually forms one bond to Carbon atom.

Hydrogen usually forms one bond Carbon atom.

Anon25 [30]3 years ago

3 0

Answer:

CHCl₃

Explanation :

In order to deduce the empirical fomula of the compound, we first express the masses given as moles:

Elements C H Cl

Masses given(g) 5.03 0.42 44.5

Then divide through by their respective molar masses:

Molar mass 12 1 35.5

(g/mol)

Number of

Moles(mol) 0.42 0.42 1.254

Dividing by

the smallest(0.42) 1 1 3

The empirical formula of the compound is

CHCl₃

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Answer:

$\large \boxed{\text{21.6 L}}$

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We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

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m/g: 24.5

(a) Moles of C₆H₁₂O₆

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(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

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3 years ago

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

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3 years ago

17. Which of the following describes a condition in which an individual would not hear an echo?

ivanzaharov [21]

The right answer is D. In a situation where the sound wave reaches the ear and the reflected wave reaches the ear less than 0.1 seconds later, the individual would not be able to hear an echo. There needs to a far more significant delay between the sound and the reflection of said sound reaching the listener's ear for the echo effect to become apparent.

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Explanation:

The given reaction equation is as follows.

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So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

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As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

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