Calculating Percentage of Stannous Fluoride in Toothpaste Sample

  • Thread starter QuarkCharmer
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In summary, the question is asking for the percentage of stannous fluoride in a 10 gram sample of toothpaste, given that it produces a 0.145 gram precipitate of stannous phosphate. The equation for calculating percentage is (actual yield)/(theoretical yield) * 100%, and the chemical equation provided is 3SnF_{2}_{(aq)} + 2K_{3}PO_{4}_{(aq)} \rightarrow Sn(PO_{4})_{2}_{(s)} + 6KF_{(aq)}. To solve, one must first convert the mass of precipitate to moles, then use a mole ratio to find the number of moles of stannous fluoride. Finally,
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QuarkCharmer
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Homework Statement


A 10.000g sample of toothpaste containing stannous fluoride gives a 0.145g precipitate of stannous phosphate. What is the percentage of SnF_2 in the toothpaste sample?

Homework Equations



[tex](Actual yield)/(Theoretical yield) * 100%[/tex]

[tex]3SnF_{2}_{(aq)} + 2K_{3}PO_{4}_{(aq)} \rightarrow Sn(PO_{4})_{2}_{(s)} + 6KF_{(aq)}[/tex]

The Attempt at a Solution



I am not positive what they are asking here. I understand that they want me to use the above formula (the percent one) to find the % composition of a certain molecule in the substance. I am thinking that they want me to first use the 10.000g to calculate the theoretical yield of the stannous fluoride using basic stoichometry and then divide the given mass of product (the precipitate) by that to get a number as a percent. Is that the correct approach?

Thanks

Edit, something is up with the LaTeX here, there should be a formula for calculating the percent (X/Y *100%) and the chemical equation, i don't know why it's rendering like it is with 2 of the chemical formulas. If you click the top equation you will see what I mean.
 
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  • #2
The question has no complications. Start working backward from the mass of precipitate. Be sure to balance the reaction and then you will have mole ratios from this.
 
  • #3
Ah, I see what you mean. The 10.000g is the total of both the reactants on the left side of the EQ. So I start with the precipitate, convert to moles, then use a mole ratio to get that in terms of the desired molecule, and convert back to grams using that mass over the 10.000 grams to form the percent equation.

For some reason I kept thinking of the 10.000g as the starting value, probably because every other problem we have done is formed that way.

Thanks alot!
 
  • #4
Most of what you say is correct. Be sure you understand that most of your 10 gram sample IS NOT THE REACTANTS; only some of it is. The question asks essentially, what proportion of the 10 grams sample of toothpaste was stannous fluoride?
 
  • #5
I get that, so when I work the product out to using the molar ratio to the questioned reactant, I would put the reactant over the whole mass of the product side. In this case over 10 grams to get the percent. Thanks
 

What is stoichiometry?

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction.

How do you solve a stoichiometry word problem?

To solve a stoichiometry word problem, you must first write out a balanced chemical equation. Then, use the given information to convert the given units to moles. Use the mole ratios from the balanced equation to determine the moles of the desired product or reactant. Finally, convert the moles back into the desired units.

What is the importance of stoichiometry in chemistry?

Stoichiometry is important in chemistry because it allows us to predict the amount of products produced from a given amount of reactants, as well as determine the amount of reactants needed to produce a desired amount of products. This is crucial in industries such as pharmaceuticals and manufacturing.

What are the key components of a stoichiometry word problem?

A stoichiometry word problem typically includes a balanced chemical equation, given information about the reactants or products (usually in the form of mass or volume), and a question about the amount of a specific product or reactant.

What are some common conversion factors used in stoichiometry?

Some common conversion factors used in stoichiometry include molar mass, Avogadro's number (6.02 x 10^23), and the mole ratio from the balanced chemical equation.

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