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Chemistry 16 Online
OpenStudy (anonymous):

Determine the mass (in g) of bromide ions in 5.99×10-3 mol of ammonium bromide.

OpenStudy (aaronq):

find the total mass: \(n=\dfrac{m}{M}\) multiply the total mass by Br's mass percent composition

OpenStudy (anonymous):

for determining the total mass, what is the mass (m)?

OpenStudy (aaronq):

that's what you're finding.

OpenStudy (aaronq):

plug in moles, and molar mass then solve for m (total mass).

OpenStudy (anonymous):

Ah!

OpenStudy (anonymous):

Mn=m (97.94g/mol)(5.99*10^-3)= 0.5867mol

OpenStudy (anonymous):

or sorry, 0.5867g

OpenStudy (aaronq):

thats good. now, find the mass percent composition of Br in the compound http://www.chem.tamu.edu/class/majors/tutorialnotefiles/percentcomp.htm

OpenStudy (anonymous):

And how would I determine Br's mass percent composition?

OpenStudy (anonymous):

oh ^^

OpenStudy (anonymous):

(79.904g/mol)/(97.94g/mol)*100= 81.5846%

OpenStudy (anonymous):

81.5846%*0.5867g= 47.8656

OpenStudy (aaronq):

when you have percents you have to convert them into decimals: 81.5846 % = 0.815846 0.815846*0.5867g=

OpenStudy (anonymous):

Right! So, 0.478656848g? (0.479g)

OpenStudy (aaronq):

yep!

OpenStudy (anonymous):

Yay!

OpenStudy (anonymous):

Thank you!!

OpenStudy (aaronq):

it feels good getting the right answer, doesn't it haha you're welcome !

OpenStudy (anonymous):

It really does!

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