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enyata [817]
3 years ago
5

A compound containing only c, h, and o, was extracted from the bark of the sassafras tree. The combustion of 43.9 mg produced 11

9 mg of co2 and 24.4 mg of h2o. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.
Chemistry
1 answer:
katen-ka-za [31]3 years ago
3 0

The compound contain only carbon, hydrogen and oxygen, let the empirical formula of compound be C_{x}H_{y}O_{z}.

The mass of compound, carbon dioxide and water is 43.9 mg, 119 mg and 24.4 mg respectively . The molar mass of compound, carbon dioxide and water is 162 g/mol, 44.01 g/mol and 18 g/mol respectively.

Converting the mass into number of moles as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Number of moles of carbon dioxide CO_{2} will be:

n=\frac{119\times 10^{-3}}{44.01 g/mol }=0.0027 mol

1 mol of CO_{2} has 1 mol of C thus, number of moles of C from 0.0027 mol of CO_{2} will be 0.0027 mol.

Molar mass of C is 12 g/mol thus, mass of C will be:

m_{C}=0.0027 mol\times 12 g/mol=0.0324 g

Number of moles of water H_{2}O will be:

n=\frac{24.4\times 10^{-3}}{18 g/mol }=0.001356 mol

1 mol of H_{2}O has 2 mol of H  thus, 0.001356 mol of H_{2}O will have 2\times 0.001356 mol=0.002712 mol mole of H.

Molar mass of H is 1 g/mol thus, mass of H will be:

m_{H}=0.002712 mol\times 1 g/mol=0.002712 g

Sum of mass of C and H will be:

m_{C+H}=0.0324+0.002712=0.035112 g

Mass of compound is 43.9 mg or 0.0439 g thus, mass of oxygen will be:

m_{O}=m_{compound}-m_{C+H}=(0.0439-0.035112)g=0.008788 g

Molar mass of oxygen is 16 g/mol thus, number of moles of oxygen will be:

n=\frac{0.008788 g}{16 g/mol}=0.00055 mol

Taking the molar ratio:

C:H:O=0.0027:0.002712:0.00055=5:5:1

Therefore, empirical formula will be C_{5}H_{5}O.

The molar mass of C_{5}H_{5}O is 81 g/mol

The molar mass of given compound is 162 that is 2 times the molar mass calculated above thus, molecular formula will be 2\times C_{5}H_{5}O=C_{10}H_{10}O_{2}.

Therefore, empirical formula and molecular formula of the compound is C_{5}H_{5}O and C_{10}H_{10}O_{2} respectively.



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Answer:

Chemical energy.

Explanation:

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5 0
3 years ago
Calculate the volume in milliliters of a 0.420mol / L barlum chlorate solution that contains 25.0 g of barium chlorate (Ba(ClO 3
Pachacha [2.7K]

<u>Answer:</u> The volume of barium chlorate is 195.65 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of barium chlorate = 25.0 g

Molar mass of barium chlorate = 304.23 g/mol

Molarity of solution = 0.420 mol/L

Volume of solution = ?

Putting values in above equation, we get:

0.420mol/L=\frac{25.0\times 1000}{304.23\times V}\\\\V=\frac{25.0\times 1000}{304.23\times 0.420}=195.65mL

Hence, the volume of barium chlorate is 195.65 mL

6 0
3 years ago
A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How
IrinaK [193]

Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 783J/^oC

c_2 = specific heat of water = 4.184J/g^oC

m_2 = mass of water = 254 g

\Delta T = change in temperature = T_2-T_1=(23.73-26.01)=-2.28^oC

Now put all the given values in the above formula, we get:

q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]

q=-4208.28J=-4.81kJ

Therefore, the amount of heat evolved by a reaction is, 4.81 kJ

7 0
3 years ago
QUESTION 2
Aleksandr [31]

Answer:

you can see the answer at the pic

4 0
3 years ago
A reaction that occurs in the internal combustion engine is n2(g) + o2(g) ⇌ 2 no(g) (a) calculate δh o and δs o for the reaction
jekas [21]
1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
7 0
2 years ago
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