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Examples: Determining Empirical & Molecular Formulas of Compounds - P2O5 & Hydrazine, Exams of Chemistry

Step-by-step solutions for determining the empirical and molecular formulas of two compounds, p2o5 and hydrazine, using given mass data and molar masses.

Typology: Exams

2021/2022

Uploaded on 09/12/2022

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laalamani 🇺🇸

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PRACTICE PROBLEMS:
1. The empirical formula for a common drying agent is P2O5. The molecule has a molar mass
of 283.88 g/mol. Find the molecular formula of the compound.
Answers:
- When determining Molecular Formulas, you must first determine the Empirical formula. In
this problem, the Empirical formula is given.
- Since you are given the Empirical formula, we can determine the Molecular formula by using
the following equation:
)(
)(
calculatemustyoumassFormulaEmpirical
problemtheingivenmassFormulaMolecular
n
- Therefore, we must first determine the Empirical formula mass for P2O5 :
P2O5 = ( 2 ) ( 30.97 g / mol ) + ( 5 ) ( 16.00 g/mol ) = 141.94 g / mol
- Lets now insert the values into the equation and solve for n:
2
94.141
88.283
mol
g
mol
g
massFormulaEmpirical
massFormulaMolecular
n
- I guess now would be a good time to discuss the significance of n”… n represents the
multiplier that we must use to determine the molecular formula. Recall, Empirical formula
represents the smallest-whole-number ratio of elements within a compound. Molecular
formula represents the actual number of elements within a compound (that is not an ionic
compound therefore, molecular (hence the name)). So
( P2O5 )n ; where n = 2
( P2O5 )2 = P4O10
- Since we calculated n to equal 2, then the Empirical formula is multiplied by a factor of two
to determine the molecular formula for the compound.
Empirical Formula = smallest whole number ratio
Molecular Formula = the actual formula for the compound
Yes, in some cases - the compounds have the same formula for empirical and molecular
pf2

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PRACTICE PROBLEMS:

1. The empirical formula for a common drying agent is P 2 O 5. The molecule has a molar mass

of 283.88 g/mol. Find the molecular formula of the compound.

Answers:

  • When determining Molecular Formulas, you must first determine the Empirical formula. In this problem, the Empirical formula is given.
  • Since you are given the Empirical formula, we can determine the Molecular formula by using the following equation:

EmpiricalFormulamass youmust calculate

MolecularFormulamass givenintheproblem

n

  • Therefore, we must first determine the Empirical formula mass for P 2 O 5 :

P 2 O 5 = ( 2 ) ( 30.97 g / mol ) + ( 5 ) ( 16.00 g/mol ) = 141.94 g / mol

  • Let s now insert the values into the equation and solve for n :

mol

g

mol

g

EmpiricalFormula mass

MolecularFormulamass

n

  • I guess now would be a good time to discuss the significance of n ”… “ n represents the multiplier that we must use to determine the molecular formula. Recall, Empirical formula represents the smallest-whole-number ratio of elements within a compound. Molecular formula represents the actual number of elements within a compound (that is not an ionic compound therefore, molecular (hence the name)). So

( P 2 O 5 )n ; where n = 2

( P 2 O 5 ) 2 = P 4 O 10

  • Since we calculated n to equal 2, then the Empirical formula is multiplied by a factor of two to determine the molecular formula for the compound.

Empirical Formula = smallest whole number ratio

Molecular Formula = the actual formula for the compound

Yes, in some cases - the compounds have the same formula for empirical and molecular

2. Hydrazine is a widely used compound. It can be used to treat waste water from chemical

plants removing ions that may be hazardous to the environment; it can be used in rocket

fuels; and it can help prevent corrosion in the pipes of electric plants. In a 32.0 gram

sample of hydrazine, there are 28.0 grams of nitrogen and 4.0 grams of hydrogen. The

molar mass of the molecule is 32.0 g/mol. What is the empirical and molecular formula

for hydrazine?

Answers:

  • Wow, what a great question Half of the question (first half) has nothing to do with solving the problem
  • Since the question is asking for both Empirical and Molecular start with determining the Empirical formula
  • Convert the grams to moles:

molN

g

g N moleN

molH

g

g H moleH

  • Determine the mole ratio:

N: 1. 00

mol

mol

H: 1. 99 2. 00

mol

mol

  • Determine the molecular formula:

Empirical Formula mass = NH 2 = ( 14.01 g/mol ) + ( 2 ) ( 1.008 g/mol ) = 16.026 g/mol

mol

g

mol

g

EmpiricalFormula mass

MolecularFormulamass

n

( NH 2 )n ; where n = 2

( NH 2 ) 2 = N 2 H 4

The ratio is:

Empirical Formula:

NH 2