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sergij07 [2.7K]
3 years ago
8

This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): An analytical chemist has det

ermined by measurements that there are moles of carbon in a sample of acetic acid. How many moles of oxygen are in the sample
Chemistry
1 answer:
jeyben [28]3 years ago
6 0

Answer:

0.054 mol O

Explanation:

<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of carbon in a sample of acetic acid. How many moles of oxygen are in the sample?</em>

<em />

Step 1: Given data

  • Chemical formula of acetic acid: CH₃CO₂H
  • Moles of carbon in the sample: 0.054 moles

Step 2: Establish the appropriate molar ratio

According to the chemical formula, the molar ratio of C to O is 2:2.

Step 3: Calculate the moles of oxygen in the sample

We will use the molar ratio to determine the moles of oxygen accompanying 0.054 moles of carbon.

0.054 mol C × (2 mol O/2 mol C) = 0.054 mol O

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True or false? an element consists of only single atoms of that element. <br> a. True <br> b. False
larisa [96]
From what i read, the answer should be  b--false
7 0
3 years ago
Which ionic compound is the most likely to dissolve in water?
lbvjy [14]
The ionic compound is CaSO4
3 0
3 years ago
The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r
Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0
3 years ago
What mass, in grams, of CO2 and H2O<br> is formed from 2.55 mol of propane?
oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of each compound:

  • C₃H₈: 44 g/mole
  • O₂: 16 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole= 44 grams
  • O₂: 5 moles* 16 g/mole= 80 grams
  • CO₂: 3 moles* 44 g/mole= 132 grams
  • H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

  • If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}

mass of CO₂= 336.6 grams

  • If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}

mass of H₂O= 183.6 grams

<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>

3 0
3 years ago
Lysine is an amino acid that is an essential part of nutrition but which is not synthesized by the human body. What is the molar
nata0808 [166]

Answer:

The molar mass of lysine using the ideal gas equation for this problem is 146.25 g/mole.

Explanation:

The ideal gas equation PV = nRT, was derived from the ABC laws (Avogadros, Boyles and Charles laws). We need to obtain the value for the number of moles n.

The parameters of this equation are:

P = 1.918 atm

V = 750.0mL = 0.75L

n = ?

R = 0.0821

T = 25 degree celcius = 25 + 273 = 298 degree kelvin.

From this formular, n = (PV)/(RT)

n = (1.918 X 0.75)/(0.0821 X 298 )

n = 0.0588

n, no of mole = mass/molar mass

0.0588 = 8.6/MM

MM = 8.6/0.0588

MM = 146.25g/mole.

4 0
3 years ago
Read 2 more answers
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