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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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2 years ago
22.4l of ammonia is reaxts with 1.406 mole of oxygen to produce NO and h2o .1.what volume of no is produced at ntp​
IceJOKER [234]

Answer:

The volume of NO is 22.4L at STP

Explanation:

Based on the reaction:

2NH3 + 5/2O2 → 2NO + 3H2O

<em>2 moles of NH3 react with 5/2 moles of O2 to produce 2 moles of NO.</em>

<em />

To solve this question, we need to find the moles of each reactant in order to find the limiting reactant as follows:

<em>Moles NH3 -Molar mass: -17.01g/mol-</em>

Using PV = nRT

PV/RT = n

<em>Where P is pressure = 1atm at STP</em>

<em>V is volume = 22.4L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 273.15K</em>

1atm*22.4L/0.082atmL/molK*273.15K = n

n = 1.00 moles of NH3

For a complete reaction of 1.00 moles of NH3 are needed:

1.00 moles NH3 * (5/2moles O2 / 2moles NH3) = 1.25 moles of O2

As there are 1.406 moles of O2, <em>the limiting reactant is NH3</em>

<em />

The moles of NO produced are the same than moles of NH3 because 2 moles of NH3 produce 2 moles of NO. The moles of NO are 1.00 moles

And as 1.00moles of gas are 22.4L at STP:

<h3>The volume of NO is 22.4L at STP</h3>

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