SOLUTION STOICHIOMETRY

SOLUTION STOICHIOMETRYBy, Sondra

What Is This? Solution Stoichiometry is the method of predicting or analyzing the volume and concentration of solutions involved in a chemical reaction.The major difference in solution stoichiometry from the general stoichiometric method is that the amount of volume and concentration are used as the conversion factors.

Calculating Stoichiometric ProblemsWrite a balanced equation, along with all quantities and conversion factors.Convert the given measurements to its chemical amount using the appropriate conversion factors.(eg. Grams, moles)Calculate the amount of the other substance using the mole ratio from your balanced equationConvert the final amount to the quantity requested.

Sample 1A student dissolves 450g of sodium chloride into 300L of water to make NaCl(H2O). What is the concentration of NaCl in the water?To solve this question we will be using 2 formulasN=M/m and C=N/V

MolesSince NaCl + H2O = NaCl(H2O) is already a balanced equation we can skip to moles.The number of moles of NaCl in a 450g sample is found as follows:22.99g/molNa+35.45g/molCl=58.44g/molNaClN=450g/58.44g/molN=7.7 moles

ConsentrationNow that we know how many moles of NaCl we have we can continue to find concentration. This is done as follows:C=7.7mol/300LC=0.026mol/LSo we now know that 450g of sodium chloride in 300L of water has a concentration of .026mol/L.

VolumeTo continue with our previous question what would the concentration of the solution be if we added 250L of water to the already 300L?To find this type of question we need the following formulaC1V1=C2V2

Missing VariableIf we put the values that we know into this equation we come up with:.026mol/L*300L=C2750LWe get 750L from adding 250L to the original 300LSolve for C2

Sidenote: In this equation you may only have one unknown variable present.

Solving.026mol/L*300L=C2750LDivide both sides by 750L(.026mol/L*300L)/750L=C2Now calculate the brackets and divide by 750LC2=.01mol/L

Sample 2Water is added to 120L of 8.50mol/L NH3 till the concentration reached 2.80mol/L. How much water is needed to reach this concentration?To complete this question we will use:C1V1=C2V2

Missing VariableFor this we need to solve for V2. Input the known variables.8.50mol/L*120L=2.80mol/L*V2

Divide both sides by 2.80mol/L and solve.

(8.50mol/L*120L)/2.80mol/L=V2V2=364L

ConclusionIn conclusion as long as you follow the steps and know those three formulas you will be able to figure out any solution question.