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Chemistry 8 Online
OpenStudy (anonymous):

Can anyone help with me with my chemistry problem, will give medal and become a fan! what is the total number of formula units in 0.89g of CsOH?

OpenStudy (cuanchi):

do you know how to calculate the molar mass of CsOH?

OpenStudy (anonymous):

no

OpenStudy (cuanchi):

do you have a periodic table?

OpenStudy (anonymous):

I can get on one the internet what am I looking for

OpenStudy (cuanchi):

atomic mas of the 3 elements in your formula and add them

OpenStudy (anonymous):

all together it's about 150 amu

OpenStudy (cuanchi):

yes that is the molecular mass of CsOH, then you calculate the number of moles in the given mass, 0.89g and then the number of formula units (molecules) with the avogadro's number

OpenStudy (anonymous):

so what is the formula of how the problem should be set up, I'm confused

OpenStudy (cuanchi):

number of moles = mass/ formula mass molecular mass (covalent compounds) = formula mass (ionic compounds)

OpenStudy (cuanchi):

do you use conversion factors?

OpenStudy (anonymous):

no

OpenStudy (cuanchi):

number of moles = mass/ molecular mass

OpenStudy (cuanchi):

then , units formula = number of moles x avogadro's number Avogadro's number= 6.02 x 10^23

OpenStudy (anonymous):

so the first part would be .89=x/150?

OpenStudy (anonymous):

is that part a solve for x kind of problem or is that something I get from the periodic table

OpenStudy (cuanchi):

no! no "x" here n= 0.89/150 = number of moles

OpenStudy (anonymous):

oh haha okay so the total answer is about 3.572

OpenStudy (cuanchi):

no

OpenStudy (cuanchi):

\[(0.98/150) \times 6.02 \times 10^{23}\]

OpenStudy (anonymous):

that's what I got from putting that into my calculator

OpenStudy (anonymous):

or is it 3.572x10^21

OpenStudy (cuanchi):

that sounds better i got 3.93 x 10 ^21

OpenStudy (anonymous):

did you do the right number, it was .89 not .98?

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