All categories

2 years ago

Calculate the empirical formula of a compound that has a composition of 5.9% (by mass) hydrogen and 94.1% (by mass) oxygen.

Chemistry

2 answers:

adelina 88 [10]2 years ago

8 0

Answer:

The empirical formula is the simplest form;

Given:

Oxygen O at 94.1% and

H at 5.9%

Assume 100grams.

94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O

5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H

There is one mole of O for each mole of H so the empirical formula is $O_1H_1$

and written as OH.

ra1l [238]2 years ago

4 0

Explanation:

$given \: that \: oxygen \: by \: 94.1\% (.941)\\ hydrogen \: by5.9\%(.059) \\ in \: 100gram \: \\ oxygen = 100 g\times .941 = 94.1 \times \frac{1(mol)}{16g} \\ = 5.88moles \: of \: oxygen \\ in \: hydrogen \: = .059\times 100 = 5.9 \times \frac{1mol}{1.002gram} \\ = 5.88mole \: of \: hydrogen \\ \: so \: here \: \: both \: oxygen \: andhydrogen = 5.88 \\their \: ratio = 1 \: 1 \\ so \: emparical \: formula = oh \\ thank \: you$

You might be interested in

Is that the right answer

julsineya [31]

Is what the right answer

8 0

2 years ago

i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i

9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

$C_8H_{18}+O_2\to CO_2+H_2O$

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

$C_8H_{18}+O_2\to8CO_2+H_2O$

Now, we balance the hydrogen:

$C_8H_{18}+O_2\to8CO_2+9H_2O$

And in the end, the oxygen:

$C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O$

We can multiply all coefficients by 2 to get integer ones:

$2C_8H_{18}+25O_2\to16CO_2+18H_2O$

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

$\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}$

For the products, we have:

$\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}$

Now, we substract the rectants from the produtcs:

$\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}$

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

$\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ$

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

$\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}$

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

$Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ$

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, m.

First, we have to heat the ice to 0 °C, so:

$\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}$

Then, we need to melt all this mass, so we use the latent heat now:

$Q_2=n\cdot6.03kJ/mol$

Converting mass to number of moles of water we have:

$\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}$

So:

$Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g$

Adding them, we have a total heat of:

$\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}$

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

$\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}$

Since we have a total of 20kg of ice, we can clculate the percent using it:

$P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%$

5 0

10 months ago

Through an uplift under the earth crust at a dirvergent boundary a _____ is formed.

postnew [5]

Out of the following given choices;

a. cliff b. fault

c. plateau d. mountain

The answer is b. A divergent boundary is a line at which two tectonic plates are moving away from each other. It is caused by the two magma convection currents in the mantle moving in opposing directions (one clockwise, the other anti-clockwise) hence dragging the crust with them. Therefore the biggest force at the boundary on the crust is that of pulling. This causes fractures and faults on the earth’s crust.

6 0

3 years ago

Acetylene gas (c2h2) reacts with oxygen gas (o2) to produce carbon dioxide (co2) and water vapor (h2o). how many liters of c2h2

MissTica

According to the balanced equation of the reaction:

2C2H2 + 5O2 → 4CO2 + 2H2O

So we can mention all as liters,

A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O

So, when we have 75L of CO2

and when we have 2 L of C2H2 reacts and gives 4 L of CO2

2C2H2 → 4CO2

∴ The volume of C2H2 required is:

= 75L / 2

= 37.5 L

B) and, when we have 75 L of CO2

and 4CO2 → 2H2O

∴ the volume of H2O required is:

= 75 L /2
= 37.5 L

C) and from the balanced equation and by the same way:

when 5 liters O2 reacts to give 4 liters of CO2

and we have 75 L of CO2:

5 O2 → 4 CO2

?? ← 75 L

∴ the volume of O2 required is:

= 75 *(5/4)

= 93.75 L

D) about the using of the number of moles the answer is:

no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.

6 0

3 years ago

Read 2 more answers

What type of reaction is shown below? a) Addition reaction b) Esterification

Gemiola [76]

Answer:

a) Addition reaction, is your answer

6 0

2 years ago

Calculate the empirical formula of a compound that has a composition of 5.9% (by mass) hydrogen and 94.1% (by mass) oxygen.​

Calculate the empirical formula of a compound that has a composition of 5.9% (by mass) hydrogen and 94.1% (by mass) oxygen.