What is the percentage by mass of ammonium phosphate in the fertilizer?

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Discussion Overview

The discussion revolves around calculating the percentage by mass of ammonium phosphate in a fertilizer sample based on a chemical reaction involving barium chloride and ammonium phosphate. The context includes a homework problem that requires understanding the relationships between the reactants and products in the reaction.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents a homework statement involving a reaction between ammonium phosphate and barium chloride, leading to the formation of barium phosphate.
  • Another participant clarifies that the percentage by mass should be calculated as the mass of ammonium phosphate divided by the total mass of the fertilizer, challenging the initial assumption about the molar mass of fertilizer.
  • Participants discuss the method of calculating moles of ammonium phosphate from the mass of barium phosphate produced in the reaction.
  • There is uncertainty expressed regarding the definition of the fertilizer in the problem, specifically whether it includes only ammonium phosphate or other components like barium chloride.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculating the percentage by mass of ammonium phosphate, but there is some uncertainty about the definition of the fertilizer and the components involved in the calculation.

Contextual Notes

There is ambiguity regarding the composition of the fertilizer and whether it solely refers to ammonium phosphate or includes other substances. Additionally, the discussion does not resolve the specifics of the calculation steps or assumptions made in the process.

littlebearrrr
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Homework Statement


Many fertilizers contain ammonium phosphate as a source of phosphorus and nitrogen. A 7.225 g sample of a fertilizer is dissolved in water and mixed with excess barium chloride solution, and the following reaction occurs:

2(NH4)3PO4(aq) + 3BaCl2(aq) -> Ba3(PO4)2(s) + 6NH4Cl(aq)

The product mixture is found to contain 3.741 g of barium phosphate. Using this info, calculate the percentage by mass of ammonium phosphate in the fertilizer.

Homework Equations



% by mass = grams of fertilizer/molar mass of fertilizer x 100
(unsure if that's correct, but that's what I think it should be)

The Attempt at a Solution



I started by taking the given amount of barium phosphate to find moles of ammonium phosphate. I think I need to use that to find grams of ammonium phosphate. However, I am also having difficulty trying to understand what the problem is specifying as the fertilizer. Is it the ammonium phosphate AND the barium chloride, or just the ammonium phosphate?
 
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littlebearrrr said:

Homework Equations



% by mass = grams of fertilizer/molar mass of fertilizer x 100
(unsure if that's correct, but that's what I think it should be)
There is no such thing as the molar mass of a fertilizer, as it is a mixture. You want to know what percentage of the total mass is due to ammonium phosphate:

% by mass = mass of ammonium phosphate / total mass of fertilizer

littlebearrrr said:

The Attempt at a Solution



I started by taking the given amount of barium phosphate to find moles of ammonium phosphate.
You can do that. You can also calculate everything using just masses.

littlebearrrr said:
I think I need to use that to find grams of ammonium phosphate. However, I am also having difficulty trying to understand what the problem is specifying as the fertilizer. Is it the ammonium phosphate AND the barium chloride, or just the ammonium phosphate?
The fertilizer is a mixture of ammonium phosphate and other stuff. The barium chloride is added in the lab.
 
Thank you DrClaude! Makes a whole lot of sense now (Answer I got: 25.65%)

Steps:
Grams of barium phosphate -> moles of barium phosphate -> moles of ammonium phosphate -> grams of ammonium phosphate.

Then divided grams of ammonium phosphate by grams of fertilizer (multiplied by 100).
 
Last edited:
littlebearrrr said:
(Answer I got: 25.65%)

Looks good.
 
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