# what's the mass of the original sample (g)?

here is a question that i don't know how it is solved :

A sample that is 75 % chloride by mass is dissolved in water and treated with an excess of AgNO3. If the mass of the AgCl precipitate that forms is 2.013 g, what was the mass of the original sample?

I hope you give me the the proper way of solving it

• Hi and welcome to Chemistry StackExchange. To best answer your question, rather than just giving the answer, it would be useful to know what part of the question you are having problems with. Could you describe what attempts you have already made, and what bits of information you think might be useful in helping answer the problem. – long Jun 19 '14 at 23:21
• thank you ! as far as i understood the problem, i was thinking that there will an equation like : xcl + agno3 --> agcl + xno3 // where x is an element // and the compound xcl is the limiting reactant because it is stated that " excess of agno3 " // therefore, i can get the moles number with the formula = mass/molarMass .. from there i will use the moles number to get xcl mass // I hope the information is clear and my "English is Ok :( " .. in short; i want the equation the describes the reaction. – Maher Jun 19 '14 at 23:34
• I have two questions to ask, which may help you. In 2.013g of AgCl, how much Chloride is there? If Chloride makes up 75% of the mass of the original sample, how much original sample was there? – long Jun 19 '14 at 23:37
• the first one: I dont know ! .. the second : the original sample will be 100% .. cl 75% and the other component 25% ?? – Maher Jun 19 '14 at 23:41

It is first of all important to recognise that all the chloride in the original sample will precipitate upon treatment with excess AgNO3. This will give you a starting point, which is a measurable amount of AgCl.

The molecular mass of AgCl = (107.87+35.45)=143.32g

Therefore, the proportion of chloride in AgCl = (35.45/143.32) = 0.2473

Therefore in 2.013g of AgCl, there is (0.2473 x 2.013) = 0.4978g of chloride.

If 0.4978g of chloride constitutes 75% of the mass of the original sample, then the amount of original sample is (0.4978 / 0.75) = 0.6637g

• I dont understand the way you used to solve this problem.. but first let me ask you: what does " treatment with or treat with " mean ? – Maher Jun 19 '14 at 23:59
• It means to "mix with" or "add to". If I treat a solution with excess AgNO3, I add, or mix an excess of AgNO3 to it. See definition 6 at [thefreedictionary.com/treat] – long Jun 20 '14 at 0:04
• ok got it .. if i ask you to write an equation to this reaction .. will you be able to write with the given information? – Maher Jun 20 '14 at 0:10
• There is no equation to write, other than the generic equation you have given in your comments above. If you make some assumptions, it is possible to guess that a molecule that is 75% chloride by weight and that is water soluble is likely to be MgCl2, but that should not form part of this question. – long Jun 20 '14 at 0:24
• Ohh yeeah !! I just got it !Thank you very muuch – Maher Jun 20 '14 at 12:01