# How many atoms are there in 320 g

### 6.4 Calculations on molar sizes

**A. Importance of the chemical equation**Example: nitrogen and hydrogen react to form ammonia

Text scheme: | Nitrogen and hydrogen | respond to | ammonia |

Formula scheme: | N | --> | 2 NH |

Raw materials | respond to | End product (s) | |

Educts | respond to | Products) | |

Number of atoms: | 2 N + 3 * 2 H atoms | = | 2 N + 2 * 3 H atoms |

8 atoms in total | = | 8 atoms in total | |

Number of particles: | 4 molecules (1 H. | <> | 2 molecules of NH |

4 mol (1 mol N | <> | 2 moles of NH | |

Masses: | 2 * 14.0067 u + 6 * 1.0079 u | = | 2 * 14.0067 u + 6 * 1.0079 u |

34.0608u total | = | 34.0608u total | |

2 * 14.0067 g + 6 * 1.0079 g | = | 2 * 14.0067 g + 6 * 1.0079 g | |

34.0608 g in total | = | 34.0608 g in total |

The chemical equation gives a lot of information about the chemical reaction.

The equation shows that 1 mol of nitrogen is required to completely react with 3 mol of hydrogen to form 2 mol of ammonia.

You can also see from the table that you need 28.0134 g (2 * 14.0067) nitrogen and 6.0474 g (6 * 1.0079) hydrogen to get 34.0608 g ammonia.

Since one can calculate all of this data, knowing the chemical equation, it is easy to do calculations on chemical reactions.

**B. Calculations**Example:

Iron burns in oxygen to form iron oxide Fe

_{3}O

_{4 }.

Calculate how many g of iron you need to get 15 g of iron oxide!

How many liters of oxygen gas do you need under standard conditions to carry out this reaction?

To solve such a task, one should proceed step by step:

a. Establishing the reaction equation

Educts: iron Fe and oxygen O_{2}

Product: iron oxide Fe_{3}O_{4}Reaction equation:

_{2}-> Fe

_{3}O

_{4}

b. Cover letter of the unknown size (by a?) And the known size:

3 Fe + 2 O_{2}-> Fe

_{3}O

_{4}? g 15 g

Explanations:

For our example we first try to find out how many g of iron it takes to get 15 g of iron oxide.

So the unknown quantity is the mass of the iron, m (Fe) =? G.

The known quantity is the mass of the iron oxide, m (Fe_{3}O_{4}) = 15 g.

This is indicated under the reaction equation.

c. Letter of the ratio of the amount of substance of the searched substance to the amount of substance of the known substance:

n (Fe) 3 -------- = - (1) n (Fe_{3}O

_{4}) 1

Explanations:

Amount of substance of the searched substance: n (Fe)

Amount of substance of the known substance: n (Fe_{3}O_{4})

The ratio of the amounts of substance n is given in the reaction equation (3 mol Fe for 1 mol Fe_{3}O_{4}).

d. Resolution according to the amount of substance of the substance sought:

n (Fe) = 3 * n (Fe_{3}O

_{4}) (2)

Explanations:

Amount of substance sought: n (Fe), because you want to calculate how much iron you need.

By multiplying (1) on both sides by n (Fe_{3}O_{4}) and simplifying one obtains the equation under d:

_{3}O

_{4}) n (Fe

_{3}O

_{4}) 1

You get:

n (Fe) * n (Fe_{3}O

_{4}) 3 * n (Fe

_{3}O

_{4}) ----------------- = ------------- n (Fe

_{3}O

_{4}) 1

Simplifying this gives equation (2):

n (Fe) = 3 * n (Fe_{3}O

_{4})

e. If one has to calculate masses, then n is replaced by m / M, but if one has to calculate volume, then n becomes by V / V_{m} replaces:

_{3}O

_{4}) ------ = ------------- (3) M (Fe) M (Fe

_{3}O

_{4})

Explanations:

You want to calculate the mass of iron, so you replace n (Fe) with m (Fe) / M (Fe).

Since the mass of Fe_{3}O_{4} is given, one replaces n (Fe_{3}O_{4}) through m (Fe_{3}O_{4}) / M (Fe_{3}O_{4}).

f. Solving for the size you are looking for:

3 * m (Fe_{3}O

_{4}) * M (Fe) m (Fe) = --------------------- (4) M (Fe

_{3}O

_{4})

Explanations:

The quantity we are looking for is the mass of the iron, m (Fe). By multiplying (3) on both sides by M (Fe) and simplifying, you get the desired result:

_{3}O

_{4}) ------ = ------------- | * M (Fe) M (Fe) M (Fe

_{3}O

_{4})

You get:

m (Fe) * M (Fe) 3 * m (Fe_{3}O

_{4}) * M (Fe) ------------- = -------------------- M (Fe) M (Fe

_{3}O

_{4})

Simplifying this gives equation (4):

3 * m (Fe_{3}O

_{4}) * M (Fe) m (Fe) = --------------------- (4) M (Fe

_{3}O

_{4})

G. Inserting the numerical values and calculation:

m (Fe_{3}O

_{4}) = 15 g (in the specification) M (Fe) = 55.847 g / mol, (Periodic Table of the Elements) M (Fe

_{3}O

_{4}) = 3 * 55.847 + 4 * 15.9994 g / mol = 231.5386 g / mol

By inserting it (the units are in brackets) you get:

3 * 15 (g) * 55.847 (g / mol) m (Fe) = ----------------------------- 231.5386 (g / mol)By calculating and simplifying you get:

m (Fe) = 10.85 gSo it takes 10.85 g of iron to make 15 g of iron oxide.

In summary, the step-by-step solution looks like this:

a. Establishing the reaction equation

3 Fe + 2 O_{2}-> Fe

_{3}O

_{4}

b. Cover letter of the unknown size (by a?) And the known size:

3 Fe + 2 O_{2}-> Fe

_{3}O

_{4}? g 15 g

c. Letter of the ratio of the amount of substance of the searched substance to the amount of substance of the known substance:

n (Fe) 3 -------- = - n (Fe_{3}O

_{4}) 1

d. Resolution according to the amount of substance of the substance you are looking for:

n (Fe) = 3 * n (Fe_{3}O

_{4})

e. If one has to calculate masses, then n is replaced by m / M, but if one has to calculate volume, then n becomes by V / V_{m} replaces:

_{3}O

_{4}) ------ = ------------- M (Fe) M (Fe

_{3}O

_{4})

f. Solving for the size you are looking for:

3 * m (Fe_{3}O

_{4}) * M (Fe) m (Fe) = --------------------- M (Fe

_{3}O

_{4})

G. Inserting the numerical values and calculation:

3 * 15 (g) * 55.847 (g / mol) m (Fe) = ----------------------------- = 10, 85 g 231.5386 (g / mol)So it takes 10.85 g of iron to make 15 g of iron oxide.

The second question can now be easily answered. Again, the general scheme is used:

a. Establishing the reaction equation

_{2}-> Fe

_{3}O

_{4}

b. Cover letter of the unknown size (by a?) And the known size:

3 Fe + 2 O_{2}-> Fe

_{3}O

_{4}? l 15 g

Explanations:

For our example you first try to find out how much l oxygen you need to get 15 g iron oxide.

So the unknown quantity is the volume of oxygen, V (O_{2}) =? l.

The known quantity is the mass of the iron oxide, m (Fe_{3}O_{4}) = 15 g.

This is indicated under the reaction equation.

c. Letter of the ratio of the amount of substance of the searched substance to the amount of substance of the known substance:

n (O_{2}) 2 -------- = - (3) n (Fe

_{3}O

_{4}) 1

Explanations:

Amount of substance of the searched substance: n (O_{2})

Amount of substance of the known substance: n (Fe_{3}O_{4})

The ratio of the amounts of substance n is given in the reaction equation (2 mol O_{2} for 1 mol Fe_{3}O_{4}).

d. Resolution according to the amount of substance of the substance sought:

n (O_{2}) = 2 * n (Fe

_{3}O

_{4}) (4)

Explanations:

Amount of substance sought: n (O_{2}), because you want to calculate what volume of oxygen you need.

By multiplying (3) on both sides by n (Fe_{3}O_{4}) and simplifying one obtains the equation under d:

_{2}) 2 -------- = - | * n (Fe

_{3}O

_{4}) n (Fe

_{3}O

_{4}) 1

You get:

n (O_{2}) * n (Fe

_{3}O

_{4}) 2 * n (Fe

_{3}O

_{4}) ----------------- = ------------- n (Fe

_{3}O

_{4}) 1

Simplifying this gives equation (2):

n (O_{2}) = 2 * n (Fe

_{3}O

_{4})

e. If one has to calculate masses, then n is replaced by m / M, but if one has to calculate volume, then n becomes by V / V_{m} replaces:

_{2}) 2 * m (Fe

_{3}O

_{4}) ------ = ------------- (5) V

_{m}(O

_{2}) M (Fe

_{3}O

_{4})

Explanations:

You want to calculate the volume of oxygen, so you replace n (O_{2}) through V (O_{2}) / V_{m}(O_{2}).

Since the mass of Fe_{3}O_{4} is given, one replaces n (Fe_{3}O_{4}) through m (Fe_{3}O_{4}) / M (Fe_{3}O_{4}).

f. Solving for the size you are looking for

2 * m (Fe_{3}O

_{4}) * V

_{m}(O

_{2}) V (O

_{2}) = ---------------------- (6) M (Fe

_{3}O

_{4})

Explanations:

The quantity we are looking for is the volume of oxygen, V (O_{2}). By multiplying (5) on both sides by V_{m} and simplify, you get the desired result:

_{2}) 2 * m (Fe

_{3}O

_{4}) ------ = ------------- | * V

_{m}(O

_{2}) V

_{m}(O

_{2}) M (Fe

_{3}O

_{4})

You get:

V (O_{2}) * V

_{m}(O

_{2}) 3 * m (Fe

_{3}O

_{4}) * V

_{m}(O

_{2}) -------------- = --------------------- V

_{m}(O

_{2}) M (Fe

_{3}O

_{4})

Simplifying this gives equation (6):

2 * m (Fe_{3}O

_{4}) * V

_{m}(O

_{2}) V (O

_{2}) = ---------------------- (6) M (Fe

_{3}O

_{4})

G. Inserting the numerical values and calculation:

m (Fe_{3}O

_{4}) = 15 g (in the specification) V

_{m}(O

_{2}) = V

_{mn}= 22.4 l / mol, since one works under standard conditions.

By inserting (the units are in brackets) you get:

2 * 15 (g) * 22.4 (l / mol) V (O_{2}) = --------------------------- 231.5386 (g / mol)

By calculating and simplifying you get:

V (O_{2}) = 2.90 l

So it takes 2.90 l of oxygen to produce 15 g of iron oxide.

In summary, the step-by-step solution looks like this:

a. Establishing the reaction equation

3 Fe + 2 O_{2}-> Fe

_{3}O

_{4}

b. Cover letter of the unknown size (by a?) And the known size:

3 Fe + 2 O_{2}-> Fe

_{3}O

_{4}? l 15 g

_{2}) 2 -------- = - n (Fe

_{3}O

_{4}) 1

d. Resolution according to the amount of substance of the substance sought:

n (O_{2}) = 2 * n (Fe

_{3}O

_{4})

_{m} replaces:

_{2}) 2 * m (Fe

_{3}O

_{4}) ------ = ------------- (5) V

_{m}M (Fe

_{3}O

_{4})

f. Solving for the size you are looking for

2 * m (Fe_{3}O

_{4}) * V

_{m}V (O

_{2}) = ------------------- (6) M (Fe

_{3}O

_{4})

G. Inserting the numerical values and calculation:

m (Fe_{3}O

_{4}) = 15 g (in the specification) V

_{m}= V

_{mn}= 22.4 l / mol, since one works under standard conditions.

By inserting (the units are in brackets) you get:

2 * 15 (g) * 22.4 (l / mol) V (O_{2}) = --------------------------- 231.5386 (g / mol)

By calculating and simplifying you get:

V (O_{2}) = 2.90 l

It takes 2.90 l of oxygen to produce 15 g of iron oxide.

The second part of the problem can also be solved by applying the law of the conservation of mass.

According to the law of the conservation of mass, one can write:

For the reaction example this means:

m (Fe) + m (O_{2}) = m (Fe

_{3}O

_{4})

By inserting the numerical values you get:

10.85 g + m (O_{2}) = 15 g

By repositioning you get:

m (O_{2}) = 15 g - 10.85 g = 4.15 g

If you know the amount of substance n (O_{2}), then one can get the volume of oxygen with the formula V = n * V_{m} to calculate. However, the equation M = m / n is also required to determine the amount of substance n (O_{2}), because the molar mass of oxygen M (O_{2}) can be calculated, and the mass of oxygen m (O_{2}) is given:

_{2}) M (O

_{2}) = ------- n (O

_{2})

By repositioning you get:

m (O_{2}) n (O

_{2}) = ------- M (O

_{2})

Since m (O_{2}) equals 4.15 g and M (O_{2}) equals 2 * 15, 9994 = 31.9988 g / mol one gets by inserting:

_{2}) = -------- ------- 31.9988 (g / mol)

By calculating you get:

n (O_{2}) = 0.130 mol

Substituting it into the equation n = V / V_{m} one can calculate the volume of oxygen:

_{2}) n (O

_{2}) = -------- V

_{m}(O

_{2})

By repositioning you get:

V (O_{2}) = n (O

_{2}) * V

_{m}(O

_{2})

Since one works under standard conditions, V_{m} equal to V_{mn} equal to 22.4 l / mol. The amount of substance n (O_{2}) was calculated to be 0.130 mol. By inserting you get:

_{2}) = 0.130 (mol) * 22.4 (l / mol)

By calculating and simplifying you get:

V (O_{2}) = 2.9 l

In principle, calculations on molar sizes should always be carried out taking into account the given scheme. Often you can achieve your goal with faster calculation methods, but the scheme is relatively foolproof.

Quantitative calculation is made easier by constant practice. You should therefore try to solve as many different tasks as possible:

6.5 Quantitative Relationships: Tasks

Back to quantitative relationships

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