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Feliz [49]
3 years ago
5

A sample of an unknown biochemical compound is found to have a percent composition of 45.46 percent carbon, 7.63 percent hydroge

n, 10.60 percent nitrogen and the balance oxygen. What is the simplest formula for this compound?
Chemistry
1 answer:
Leona [35]3 years ago
3 0

Answer:

Formular = C₅H₁₁NO₃

Explanation:

The empirical formular is the simplest formular of a compound can have.

We use the steps below to obtain the empirical formular;

Step 1: Obtain the mass of each element present in grams. Element % = mass in g = m.

Carbon = 45.46% = 45.46g

Hydrogen = 7.63% = 7.63g

Nitrogen = 10% = 10g

Oxygen = 100% - (45.46% + 7.63% + 10%) = 36.31% = 36.31g

Step 2: Determine the number of moles of each type of atom present.

Molar amount (M) = m/atomic mass

Carbon = 45.46 / 12 = 3.7883

Hydrogen = 7.63 / 1 = 7.63

Nitrogen = 10 / 14 = 0.7143

Oxygen = 36.91 / 16 = 2.3069

Step 3: Divide the number of moles of each element by the smallest number of moles. Smallest = 0.7143

Carbon = 3.7883 / 0.7143 = 5.3035

Hydrogen = 7.63 / 0.7143 = 10.67

Nitrogen =  0.7143 / 0.7143 = 1

Oxygen = 2.2693 / 0.7143 = 3.1770

Step 4: Convert numbers to whole numbers

Carbon = 5

Hydrogen = 11

Nitrogen = 1

Oxygen = 3

Formular = C₅H₁₁NO₃

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1. watershed

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5 0
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Read 2 more answers
(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum
ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

4 0
1 year ago
It takes 11.2 kj of energy to raise the temperature of 145 g of benzene from 22.0°c to 67.0°c. what is the specific heat of benz
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We can use the heat equation,
Q = mcΔT 

where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
</span>c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
           c = 1.72 J g⁻¹ °C⁻¹

Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
7 0
3 years ago
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