A compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% 0. Determine the empirical formula and the molecular formula.

A) CHO and C6H6O6

B) C3H3O and C6H6O2

C) C3HO and C6H2O2

D) CH2O and C4H8O4

E) CH4O and C3H12O3

Answers

Answer 1

Answer:

B) C3H3O and C6H6O2

Explanation:

Given data:

Molar mass of compound = 100 g/mol

Percentage of hydrogen = 5.45%

Percentage of carbon = 65.45%

Percentage of oxygen = 29.09%

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 5.45 / 1.01 = 5.4

Number of gram atoms of O = 29.09/ 16 = 1.8

Number of gram atoms of C = 65.45 / 12 = 5.5

Atomic ratio:

            C                      :      H            :         O

           5.5/1.8              :     5.4/1.8     :        1.8/1.8

            3                      :        3          :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 12×3 + 1.01 ×3 + 16×1 = 55.03  

n = 100 / 5503

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₃H₃O)

Molecular formula = C₆H₆O₂


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->
3 CO2 + 4H2O

Answers

Answer:

120 g CO₂

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional AnalysisExplanation:

Step 1: Define

[RxN - Balanced] C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

[Given] 150 g O₂

Step 2: Identify Conversions

[RxN] 5 mol O₂ → 3 mol CO₂

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mass of C - 12.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Stoich

[DA] Set up:                                                                                                     [tex]\displaystyle 150 \ g \ O_2(\frac{1 \ mol \ O_2}{32.00 \ g \ O_2})(\frac{3 \ mol \ CO_2}{5 \ mol \ O_2})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})[/tex][DA] Multiply/Divide [Cancel out units]:                                                         [tex]\displaystyle 123.778 \ g \ CO_2[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

123.778 g CO₂ ≈ 120 g CO₂

What is the percent composition of the element hydrogen in the compound methane CH4

Answers

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Answers

Answer:

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4.75/0.50

This leaves you with 9.5 m


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Hope this is clear! Have a nice day.

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Answer:

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Answers

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General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Atomic Structure

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Determine the amount of grams of N2 is produced If 4.03 moles of NH3 react?

Answers

Your answer and workings are all mentioned in the picture below.

Taking into account the reaction stoichiometry, 56.42 grams of N₂ are formed when 4 moles of NH₃ react.

Reaction stoichiometry

In first place, the balanced reaction is:

4 NH₃ + 3 O₂  → 2 N₂+ 6 H₂O

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

NH₃: 4 molesO₂: 3 molesN₂: 2 molesH₂O: 6 moles

The molar mass of the compounds is:

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

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Mass of N₂ formed

The following rule of three can be applied: if by reaction stoichiometry 4 moles of NH₃ form 56 grams of N₂, 4.03 moles of NH₃ form how much mass of N₂?

[tex]mass of N_{2} =\frac{4.03 moles NH_{3} x56 grams of N_{2} }{4 moles NH_{3}}[/tex]

mass of N₂= 56.42 grams

Then, 56.42 grams of N₂ are formed when 4 moles of NH₃ react.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ2

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Answers

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Answers

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Answers

Answer:

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Explanation:

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How many moles of NH3 could you make with 4.2021 g of N2?



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Answers

Moles of NH3 = 0.3 moles

Further explanationGiven

N2 + 3H2 → 2NH3

4.2021 g of N2

Required

moles of NH3

Solution

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= 4.2021 g : 28 g/mol

= 0.15

From the equation, mol NH3 :

= 2/1 x mol N2

= 2/1 x 0.15

= 0.3 moles

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Answers

CH4 : Mr. 12+(1x4) =16
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