The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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According to the periodic table, carbon's molar mass is 12.011 grams per mole (that's the small number under the element). So, just multiply like this to get the answer:
So, there are approximately 0.208 grams in 2.5 moles of carbon.
Answer:
The concentration resulting solution = 0.350 M
Explanation:
In case of dilution , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = initial concentration ,
V₁ = initial volume ,
M₂ = final concentration , i.e. , concentration after dilution ,
V₂ = final volume .
from , the question ,
M₁ = 0.85 M
V₁ = 4.12 L
M₂ = ?
V₂ = 10.0 L
Using the above formula , the molarity of the final solution after dilution , can be calculated as ,
M₁V₁ = M₂V₂
0.85 M * 4.12 L = M₂ * 10.0 L
M₂ = 0.85 M * 4.12 L / 10.0 L
M₂ = 0.350 M
Answer:
1577 °C
Explanation:
Step 1: Calculate the volume of the cylinder
The diameter of the base (d) is 13.0 cm and the height (h) is 15.6 cm. We will use the following expression.
V = h × π × (d/2)²
V = 15.6 cm × π × (13.0 cm/2)² = 8.28 × 10³ cm³ = 8.28 × 10³ mL = 8.28 L
Step 2: Calculate the moles of BF₃
We have 0.0985 kg (98.5 g) of BF₃, whose molar mass is 67.81 g/mol.
98.5 g × 1 mol/67.81 g = 1.45 mol
Step 3: Convert 2.70 MPa to atm
We will use the conversion factor 1 MPa = 9.86923 atm.
2.70 MPa × 9.86923 atm/1 MPa = 26.6 atm
Step 4: Calculate the maximum safe operating temperature
We will use the ideal gas equation.
P × V = n × R × T
T = P × V / n × R
T = 26.6 atm × 8.28 L / 1.45 mol × (0.0821 atm.L/mol.K) = 1850 K
We can convert 1850 K to Celsius using the following expression.
°C = K - 273.15 = 1850 K - 273.15 = 1577 °C
Its a *hydrocarbon*, and it's functional group is *Propane*. c3h8
