Chemistry, 14.01.2021 16:20, breannagrd4075

Determine the correct empirical formula of a compound containing 26.68% carbon, 2.24% hydrogen, and 71.08% oxygen.

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answered: Obeilskzexal

The correct empirical formula of a compound containing 26.68% carbon, 2.24% hydrogen, and 71.08% oxygen is CHO₂

The empirical formulae shows us how elements in a compound are in their simplest ratio.

From the question, we can see

Carbon (C) with 26.68%

Hydrogen (H)with 2.24%

Oxygen (O) with 71.08%

Now, the empirical formula of the compound can begotten following these steps:

We find the molar amounts of carbon, hydrogen and oxygen in a 100-g sample by dividing each element’s mass by its molar mass

Carbon = 26.68 /12 = 2.223

Hydrogen= 2.24 / 1 = 2.24

Oxygen = 71.08 / 16 = 4.4425

We divide each molar amount by the least of the three

C arbon = 2.223 / 2.223 = 1

H ydrogen= 2.24 / 2.223 = 1

Oxygen = 4.4425 /2.223 = 2

Giving us the empirical formula of the compound as CHO₂

See related question here:link

answered: matthewmalte7329

CHO₂

Explanation:

The following data were obtained from the question:

Carbon (C) = 26.68%

Hydrogen (H) = 2.24%

Oxygen (O) = 71.08%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 26.68%

H = 2.24%

O = 71.08%

Divide by their molar mass.

C = 26.68 /12 = 2.223

H = 2.24 / 1 = 2.24

O = 71.08 / 16 = 4.4425

Divide by the smallest.

C = 2.223 / 2.223 = 1

H = 2.24 / 2.223 = 1

O = 4.4425 /2.223 = 2

Thus the empirical formula of the compound is CHO₂

answered: Guest

in scientific notation, the value of the speed of light in a vacuum

will be written as 2.995 * 10^8. when this value is represented in the form of

a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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Determine the correct empirical formula of a compound containing 26.68% carbon, 2.24% hydrogen, and...