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zalisa [80]
3 years ago
12

A sample in the laboratory is found to contain 3.36 grams of hydrogen, 20.00 grams of carbon, and 26.64 grams of oxygen. The mol

ecular mass is 180.156 g/mol. Determine both the empirical formula and the molecular formula.
Chemistry
1 answer:
KonstantinChe [14]3 years ago
5 0

Answer:

Empirical formula is CH₂O.

Molecular formula = C₆H₁₂O₆

Explanation:

Given data:

Mass of hydrogen = 3.36 g

Mass of carbon = 20.00 g

Mass of oxygen = 26.64 g

Molar mass of compound = 180.156 g/mol

Empirical formula = ?

Molecular formula = ?

Solution:

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number

Number of gram atoms of H = 3.36 / 1.01 = 3.3

Number of gram atoms of O = 26.64 / 16 = 1.7

Number of gram atoms of C = 20 / 12 = 1.7

Atomic ratio:

            C                      :        H            :         O

           1.7/1.7                :     3.3/1.7       :       1.7/1.7

              1                     :           2          :        1

C : H : O = 1 : 2 : 1

Empirical formula is CH₂O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = CH₂O = 12×1 + 2× + 16

Empirical formula mass = 30

n = 180.156 / 30

n = 6

Molecular formula = n (empirical formula)

Molecular formula = 6 (CH₂O)

Molecular formula = C₆H₁₂O₆

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KonstantinChe [14]

Answer:

Gastropoda.

Explanation:

3 0
3 years ago
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A gas starts at a volume of 23 L and a pressure of 1.23 atm. What is the new pressure if you
Anna71 [15]

Answer:

<h2>1.89 atm</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{1.23 \times 23}{15}  =  \frac{28.29}{15}  \\  = 1.886

We have the final answer as

<h3>1.89 atm</h3>

Hope this helps you

7 0
3 years ago
Please help *chemistry*
Daniel [21]
3 because I just did it
8 0
3 years ago
Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1
timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺  + 1en → [Ni(H₂O)₄(en)]²⁺  ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺  + 1en → [Ni(H₂O)₂(en)₂]²⁺  ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en →  [Ni(en)₃]²⁺  ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺  + 3en → [Ni(en)₃]²⁺  ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ =  - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0
3 years ago
How many moles of propane<br> react when 294 g of CO2 form?<br><br> C3H8 +502 → 3CO₂ + 4H₂O
tatiyna

2.23 moles of propane react when 294 g of CO₂ is formed .

<h3>What is moles ?</h3>

Moles is a unit which is equal to the molar mass of an element.

A reaction is given

C₃H₈ +50₂ → 3CO₂ + 4H₂O

Grams of CO₂ formed = 294 gm

In moles = 294 /44 = 6.68 moles.

Let x be the moles of C₃H₈ is x

Mole ratio of CO₂ to C₃H₈ = 3 : 1

so

6.68 /x = 3/1

x = 6.68 /3 = 2.23 moles

Therefore 2.23 moles of propane react when 294 g of CO₂ is formed .

To know more about Moles

brainly.com/question/26416088

#SPJ1

7 0
2 years ago
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