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Vinil7 [7]
3 years ago
8

A 4.50 g sample is analyzed and found to contain 3.22 g of carbon, 0.300 g of hydrogen, 0.221 g of nitrogen, and the rest is oxy

gen. The molar mass of the actual compound is found to be 285 g/mol. Find the empirical formula and then the molecular formula.
Chemistry
1 answer:
Kitty [74]3 years ago
8 0

The empirical formula and the molecular formula are the same : C₁₇H₁₉NO₃

<h3>Further explanation</h3>

Given

3.22 g of carbon, 0.300 g of hydrogen, 0.221 g of nitrogen, and the rest is oxygen.

Required

The empirical formula and then the molecular formula.

Solution

Mass of Oxygen = 4.5 - (3.22+0.3+0.221)=0.759 g

Mol ratio of elements C : H : N : O =

3.22/12 : 0.3/1 : 0.221/14 : 0.759/16 = 0.268 : 0.3 : 0.016 : 0.047

Divide by 0.016 :

16.75 : 18.75 : 1 : 2.9375 = 17 : 19 : 1 : 3

(C₁₇H₁₉NO₃)n = 285

(12.17+19+14+3.16)n = 285

(285)n = 285

n = 1

Then the empirical formula and the molecular formula are the same

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Answer: 59 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2H_2O\rightarrow 2H_2+O_2

Given: mass of hydrogen = 6.6 g

mass of oxygen = 52.4 g

Mass of products = Mass of hydrogen + mass of oxygen = 6.6 +52.4 = 59 g grams

Thus mass or reactant = mass of water

Mass of reactants = mass of products = 59 g

Thus the mass of water initially present was 59 g.

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4 years ago
What is the brown liquid for that the grasshopper may “spit” out?
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Tobacco juice would be your correct answer.
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Consider the motion diagram. It illustrates a car's velocity (V) and acceleration (A). The BEST description of a car’s velocity
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What is the pH of a 1 ´ 10-5 M KOH solution?
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pOH = - log [ OH⁻ ]

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3 0
3 years ago
2. A 10.0 g ice cube, initially at 0.0ºC, is melted in 100.0 g of water that was initially 20.0ºC. After the ice has melted, the
arlik [135]

Answer:

The heat lost  by the water  Q_{water}  = 3.8 KJ

The heat gain by ice  Q_{ice} = 228.76 J

The heat required to melt the ice Q_{melt} = 3340 J

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Initial temperature of ice cube T_{ice} = 0 °c

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Initial temperature of water T_{w} = 20 °c

Final temperature of mixture T_{f} = 10.93 °c

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⇒ Q_{water}  = 3.8 KJ

This is the heat lost  by the water.

(b). Heat gained by the ice cube Q_{ice} = m_{ice} C_{ice} (T_{f} - T_{ice}  )

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