Calculating Percentage of Stannous Fluoride in Toothpaste Sample

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SUMMARY

The discussion focuses on calculating the percentage of stannous fluoride (SnF2) in a toothpaste sample weighing 10.000g, which yields a precipitate of stannous phosphate (Sn(PO4)2) weighing 0.145g. The correct approach involves using stoichiometry to determine the theoretical yield of SnF2 based on the mass of the precipitate and the balanced chemical equation: 3SnF2 + 2K3PO4 → Sn(PO4)2 + 6KF. The final percentage is calculated by dividing the mass of SnF2 by the total sample mass and multiplying by 100%. The key takeaway is understanding the mole ratios and the distinction between the total sample mass and the mass of the reactants.

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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



(Actual yield)/(Theoretical yield) * 100%

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

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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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.
 
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 a lot!
 
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?
 
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
 

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