Calculating Molarity of FeSCN with Given Moles and Known Reactants

In summary, by combining 5 mL of 0.20 M Fe with 1 mL of 4.0x10^-4 M HSCN and 4 mL of 0.5 M HNO3, we are asked to find the molarity of FeSCN. According to the given statement, the concentration of Fe is high enough that we can assume all SCN is in the complexed form. We can calculate the moles of HSCN by multiplying the molarity by the amount added, which is equivalent to the moles of FeSCN formed. To find the molarity of FeSCN, we need to know the total volume of the mixture. In this case, the total volume is 10
  • #1
Prone17
7
0
i combined 5 mL of 0.20 M Fe with 1 mL of 4.0x10^-4 M HSCN and 4 mL of 0.5 M HNO3

i am asked to find the molarity of FeSCN.

i am given this statement: "The concentration of Fe is so high relative to that of HSCN that we can assume that all SCN present is in the complexed form, FeSCN. In other words, HSCN is the liming reagent and the number of moles of HSCN added equals the number of moles of FeSCN that forms."

so i calculated the moles of HSCN by multiplying the molarity by the amount added -> (4.0x10^-4)(0.001) = 4.0x10^-7
apparently this is the same as the moles of FeSCN.

how do i find the molarity of FeSCN? i know the moles, so all i need is the volume (Liters), but i don't know how much to use in the equation... any suggestions?

Thanks
 
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  • #2
You have mixed everything together, right? So what is the final volume of the mixture?
 
  • #3
A matter of simple accounting for total volume of the prepared solution of FeSCN. Sum of the volumes used in preparing the solution.
 
  • #4
the total volume would be 10 mL

thanks guys, i got it from here (hopefully) :D
 
  • #5
for your question. To find the molarity of FeSCN, you can use the formula M = moles of solute / volume of solution (in liters). In this case, you have already calculated the moles of FeSCN to be 4.0x10^-7. To find the volume of the solution, you can add up the volumes of all the reactants used (5 mL + 1 mL + 4 mL = 10 mL = 0.01 L). So, the molarity of FeSCN would be 4.0x10^-7 moles / 0.01 L = 4.0x10^-5 M.

It is important to note that the assumption made in the given statement is crucial in this calculation. If the concentration of Fe was not significantly higher than that of HSCN, then the assumption that all SCN is in the complexed form may not hold true and the molarity of FeSCN would need to be calculated using a different approach.
 

1. How do I calculate the molarity of FeSCN with given moles and known reactants?

To calculate the molarity of FeSCN, you will need to use the formula: Molarity (M) = moles of solute (mol) / volume of solution (L). First, determine the number of moles of FeSCN present in the solution by using the given amount of FeSCN and its molar mass. Then, divide the number of moles by the volume of the solution, which is typically given in liters (L). This will give you the molarity of FeSCN in the solution.

2. What are the units for molarity?

Molarity is expressed as moles per liter (mol/L) or moles per cubic decimeter (mol/dm^3). These units represent the concentration of a solute in a given volume of solution.

3. Can I use any unit for volume when calculating molarity?

No, the volume used in the molarity formula must be in liters (L) to get the correct units for molarity. If the volume is given in a different unit, such as milliliters (mL), it must be converted to liters by dividing by 1000.

4. How does the number of moles of reactants affect the molarity of FeSCN?

The number of moles of reactants used in a chemical reaction directly affects the amount of FeSCN produced and, therefore, the molarity of FeSCN in the solution. The more moles of reactants used, the more moles of FeSCN will be produced, resulting in a higher molarity.

5. Why is it important to know the molarity of FeSCN in a solution?

The molarity of FeSCN is important because it represents the concentration of FeSCN in the solution. This information is necessary for accurately determining the amount of FeSCN present in a given volume of solution, which can be useful in various scientific experiments and applications.

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