**Dougherty Valley High School Chemistry**

**Stoichiometry Online Help**

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Problems Here:**

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**What is Stoichiometry?**

The word **stoichiometry**
derives from two Greek words: *stoicheion* (meaning "element")
and *metron* (meaning "measure"). Stoichiometry deals with
calculations about the masses (sometimes volumes) of reactants and products
involved in a chemical reaction. It usually determines how much of a substance
is __expected__ to react or be produced in a chemical reaction. It is a very
mathematical part of chemistry, so be prepared for lots of calculator use.

Jeremias
Benjaim Richter (1762-1807) was the first to describe the principles of
stoichiometry. In 1792 he wrote:

"Die *stöchyometrie*
(Stöchyometria) ist die Wissenschaft die quantitativen oder Massenverhältnisse
zu messen, in welchen die chymischen Elemente gegen einander stehen."
(German translated into English: Stoichiometry is the science of measuring the
quantitative proportions or mass ratios in which chemical elements stand to one
another.)

**The key steps (in order) to doing stoichiometry:**

1. Balance your equation
first.

2. Convert from grams to
moles using molar mass.

3. Determine the
limiting reagent [if necessary] (Use mole ratios to figure out.)

4. Use ratios to find
the moles of the reactant or product you need to find.

5. Convert from moles
back to grams of the new substance using that substance's molar mass.

6. Calculate percent
yield [if necessary] (Divide actual yield by theoretical yield X 100%)

**Sample Example Problem [to show you how it's done]:**

**Problem:** How many grams of silicon carbide are produced
when 50.0 g of silicon dioxide is heated with 32.0 g of carbon?

SiO_{2} (s)
+ C (s) ā SiC (s)
+ CO (g) [unbalanced]

**1.**** Balance your equation first.**

SiO_{2} (s)
+ **3**C (s) ā SiC
(s) +
**2**CO (g) [balanced]

**2.**** Convert from grams to moles using
molar mass.**

To get moles from grams of silicon dioxide (SiO_{2}): [molar mass = 60 g/mol]

50.0 g SiO_{2} **X**
( 1 mol / 60 g ) = 0.833 mol SiO_{2}

To get moles from grams of carbon (C): [molar mass = 12 g/mol]

32.0 g C
**X** ( 1 mol / 12
g ) = 2.667 mol C

**3.**** Determine the limiting reagent [if
necessary] (Use mole ratios to figure out.)**

Figure out how many moles of C would need to react with
0.833 moles SiO_{2}:

* From your balanced equation, the ratio is 3
mol C: 1 mol SiO_{2}.

0.833 mol SiO_{2} **X**
( 3 mol C / 1 mol SiO_{2} )
= 2.5 mol C needed

* But you have 2.667 mol C to start with and
need 2.5 mol C to react, so you have extra C and limiting reagent is SiO_{2}

* To calculate excess: 2.667 - 2.500 = 0.167
mol C in excess

OR

Figure out how many moles of SiO_{2} would need
to react with 0.833 moles C:

* From your balanced equation, the ratio is 1
mol SiO_{2} : 3 mol C.

2.667 mol C
**X** ( 1 mol SiO_{2}
/ 3 mol C ) = 0.889 mol SiO_{2}
needed

*
But you have 0.833 mol SiO_{2} to start with and need 0.889 mol SiO_{2}
to react, so you have SiO_{2 }limiting reagent because you don't have
enough SiO_{2 }needed to react

* Compare 0.889 mol SiO_{2 }needed vs.
0.833 mol SiO_{2 }you started with

**4.**** Use ratios to find the moles of
the reactant or product you need to find.**

Since
SiO_{2} is the limiting reagent, you use the moles of SiO_{2 }if
you have to calculate how many grams of silicon carbide (the product you're
trying to find) are being produced. Use your mole ratios from your balanced
equation:

0.833 mol SiO_{2} **X**
( 1 mol SiC / 1 mol SiO_{2} )
= 0.833 mol SiC produced

**5.**** Convert from moles back to grams
of the new substance using that substance's molar mass.**

To get grams from moles of carbon (SiC): [molar mass = 40 g/mol]

0.833 mol SiC X ( 40 g / 1
mol ) = **33.33 g SiC produced**

**Answer
to Problem:** **33.33 g SiC
produced = the theoretical yield**