# What is a stoechiometric equation

## Stoichiometric calculations

In order to be able to carry out the stoichiometric calculations, we should know the following chemical quantities and their units:

Size with unity | description | description |
---|---|---|

$ m \ [g] $ | Dimensions | Weight of a substance |

$ M \ \ left [\ frac {g} {mol} \ right] $ | Molar mass | Proportionality factor between mass and amount of substance |

$ n \ [mol] $ | Amount of substance | mol is the amount of substance in a portion of substance |

$ N_A = 6 {,} 022 \ cdot 10 ^ {23} \ \ left [\ frac {1} {[mol]} \ right] $ | Avogadro's constant | Constant: number of particles per amount of substance |

$ N $ | Particle number | Number of particles in a portion of the substance |

$ V \ [L] $ | volume | |

$ V_m = 22 {,} 4 \ \ left [\ frac {{L}} {{mol}} \ right] $ | Molar volume | Volume occupied by 1 $ [mol] $ of the substance; under normal conditions $ V_m = 22 {,} 4 \ \ left [\ frac {{L}} {{mol}} \ right] $ |

$ c \ \ left [\ frac {{mol}} {{L}} \ right] $ | concentration | Amount of substance per volume |

**The most important stoichiometric formulas:**

\ begin {align *}

\ text {1.} \ quad n = N \ cdot N_A \ quad \ quad

\ text {2.} \ quad M = \ frac {m} {n} \ quad \ quad

\ text {3.} \ quad V_m = \ frac {V (\ text {Gas})} {n (\ text {Gas})} \ quad \ quad

\ text {4.} \ quad c = \ frac {n} {V}

\ end {align *}

A math problem in chemistry usually includes the following steps:

- Establish reaction equation
- Establish a molar ratio
- Conversion of the known size into the amount of substance
- Calculation of the amount of substance of the size you are looking for
- Calculate the size you are looking for from the amount of material

Example: Iron and oxygen react to form 10 g of iron (III) oxide. Enter the mass of iron used and the volume of oxygen consumed.

**1. Set up the reaction equation:** \ begin {align *} {4Fe + 3O_2 -> 2Fe_2O_3} \ end {align *}

**2. Establish a molar ratio**

We always establish the ratio of the amount of substance from the amount of substance of which a size is sought and the amount of substance of the substance of which a size is given.

So here the molar ratio from the molar amount of iron and iron (III) oxide and the molar ratio from oxygen and iron (III) oxide. A little tip: If we always write the amount of substance of the substance we are looking for in the counter, it will be easier to solve for this amount of substance later.

\ begin {align *}

\ frac {n ({Fe})} {n ({Fe_2O_3})} = \ frac {4} {2} = \ frac {2} {1} = 2 \ quad \ text {and} \ quad \ frac { n ({O_2})} {n ({Fe_2O_3})} = \ frac {3} {2} = 1 {,} 5

\ end {align *}

**3. Conversion of the known size into the amount of substance**

In our example, the mass of iron (III) oxide is given (m = 10 g). The formulas, in which both substance amount and mass occur, is: \ begin {align *} M = \ frac {m} {n} \ end {align *}

In order to be able to calculate the amount of substance, we also need the molar mass $ M $. To do this, we take a look at the periodic table. The molar mass of iron is 55 {,} 85 [g] / [mol], that of oxygen 16 \ [g] [mol]. Two iron atoms and three oxygen atoms are bound in iron (III) oxide. To calculate the molar mass of ferric oxide, we add twice the molar mass of iron and three times the molar mass of oxygen.

\ begin {align *}

M ({Fe_2O_3}) = 2 \ times M ({Fe}) + 3 \ times M ({O}) = 2 \ times 55 {,} 85 \ \ frac {{g}} {{mol}} + 3 \ cdot 16 \ \ frac {{g}} {{mol}} = 159 {,} 70 \ \ frac {{g}} {{mol}}

\ end {align *}

Now we know two quantities from the formula and calculate the amount of substance n.

\ begin {array} {crcll}

& M & = & \ frac {m} {n} & | \ cdot n \

\ Leftrightarrow & M \ cdot n & = & m & |: M \

\ Leftrightarrow & n & = & \ frac {m} {M} &

\ end {array}

We can now insert the mass and the molar mass into the formula solved for the amount of substance:

\ begin {align *}

n = \ frac {10 \ {g}} {159 {,} 70 \ \ frac {{g}} {{mol}}} = 0 {,} 0626 \ {mol}

\ end {align *}

**4. Calculation of the amount of substance of the substance sought**

In the second step, we have already established the required proportions of the amount of substance. We now resolve this according to the amount of substance of the substance we are looking for and use the amount of iron (III) oxide calculated in step three.

\ begin {align *}

\ begin {array} {crcl}

& \ frac {n {Fe}} {n ({Fe_2O_3})} & = & 2 \ quad \ quad | \ cdot n {Fe_2O_3} \

\ Leftrightarrow & n {Fe} & = & 2 \ cdot n ({Fe_2O_3}) = 2 \ cdot 0 {,} 0626 \ [mol] = 0 {,} 1252 \ [mol] \ \

& \ frac {n {O_2}} {n {Fe_2O_3}} & = & 1 {, 5} \ quad | \ cdot n ({Fe_2O_3}) \

\ Leftrightarrow & n {O2} & = & 1 {,} 5 \ times n {Fe_2O_3} = 1 {,} 5 \ times 0 {,} 0626 \ [mol] = 0 {,} 0939 \ [mol]

\ end {array}

\ end {align *}

5. Calculate the required size from the amount of substance To calculate the mass of the iron used, we use the formula M = m = n again. In this case we rearrange it according to the mass m:

\ begin {align *}

\ begin {array} {crcl}

& M & = & \ frac {m} {n} \ quad \ quad | \ cdot n \

\ Leftrightarrow & m & = & M \ cdot n = 55 {,} 85 \ \ frac {g} {mol} \ cdot 0 {,} 1252 \ {mol} = 6 {,} 99 \ {g}

\ end {array}

\ end {align *}

In the case of oxygen, the volume consumed is required. The formula, which contains both the amount of substance and the volume, is

\ begin {align *}

V_m = \ frac {V} {n}

\ end {align *}

The molar volume is always 22.4 L = mol. So we solve the formula for the volume and then only insert the amount of substance and molar volume.

\ begin {align *}

\ begin {array} {crcl}

& V_m & = & \ frac {V} {n} \ quad \ quad | \ cdot n \

\ Leftrightarrow & V & = & V_m \ cdot n = 22 {,} 4 \ \ frac {L} {mol} \ cdot 0 {,} 0939 \ {mol} = 2 {,} 1 \ {L}

\ end {array}

\ end {align *}

The most important parameter in chemistry is the amount of substance n. This is mainly due to the fact that only these are comparable with each other and thus the ratio of the amount of substance must be determined in almost all stoichiometric calculations.

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