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Ksenya-84 [330]
4 years ago
14

A laboratory analysis of vanillin, the flavoring agent in vanilla, indicates that its mass percent composition is 63.15% C, 5.30

% H, and 31.55% O. Determine the empirical formula of vanillin.
Chemistry
1 answer:
Nesterboy [21]4 years ago
6 0

Answer:

C3H3O

Explanation:

Now, what we need to do is to divide the percentage compositions by the relative atomic masses.

The relative atomic mass of carbon 12, the relative atomic mass of hydrogen is 1 while the atomic mass of oxygen is 16.

We now do the divisions as follows:

Carbon = 63.15/12 = 5.2625

Oxygen = 31.55/16 = 1.971875

Hydrogen = 5.30/1 = 5.30

Now,we now divide by the smallest I.e that of oxygen.

Carbon = 5.2625/1.971875 = 2.668

Oxygen = 1.971875/1.971875 = 1

Hydrogen = 5.3/1.971875= 2.688

We can round up the ratio of carbon and hydrogen to 3.

Thus, the empirical formula is C3H3O

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A molecule is the ___________particle of a substance that can normally exist independenty
Goryan [66]

Answer:

a molecule is the <u>smallest</u> particle of a substance that can normally exist independenty

5 0
2 years ago
A balloon contains 0.950 mol of nitrogen gas and has a volume of 25.5 L. How many grams of N2 should be released from the balloo
lianna [129]

Answer:

Mass released = 8.6 g

Explanation:

Given data:

Initial number of moles nitrogen= 0.950 mol

Initial volume = 25.5 L

Final mass of nitrogen released  = ?

Final volume = 17.3 L

Solution:

Formula:

V₁/n₁  = V₂/n₂

25.5 L / 0.950 mol = 17.3 L/n₂

n₂ =  17.3 L× 0.950 mol/25.5 L

n₂ = 16.435 L.mol /25.5 L

n₂ = 0.644 mol

Initial mass of nitrogen:

Mass = number of moles × molar mass

Mass = 0.950 mol × 28 g/mol

Mass = 26.6 g

Final mass of nitrogen:

Mass = number of moles × molar mass

Mass = 0.644 mol × 28 g/mol

Mass = 18.0 g

Mass released = initial mass - final mass

Mass released = 26.6 g - 18.0 g

Mass released = 8.6 g

6 0
3 years ago
The majority of elements found in the periodic table can be classified as what
Karo-lina-s [1.5K]
They can be classified as metals.
4 0
3 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 19.2 g of carbon were burned in the presence of
SVETLANKA909090 [29]
70.4g of carbon dioxide was produced
5 0
4 years ago
How many molecules are in 1.5 moles of carbon dioxide
Karo-lina-s [1.5K]

Answer:

<h2>9.03 × 10²³ molecules</h2>

Explanation:

The number of molecules can be found by using the formula

N = n × L

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 1.5 × 6.02 × 10²³

We have the final answer as

<h3>9.03 × 10²³ molecules</h3>

Hope this helps you

3 0
3 years ago
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