MHB Calculate Oxygen Mass at RT & Pressure: 78 cm3

[markosheehan]

Calculate the mass of 78 cm3 of oxygen at room temperature and pressure .

Im trying to work out the number of moles and then multiply by 32.

 

[Joppy]

Calculate the mass of 78 cm3 of oxygen at room temperature and pressure .

Im trying to work out the number of moles and then multiply by 32.

$n(O) = \dfrac{V(O)}{V_m(O)}$

$m(O) = n(O)M(O)$.

$V_m$ is the molar gas volume.

 

[markosheehan]

Im sorry I am still confused. I know the right answer is .104g

 

[I like Serena]

Im sorry I am still confused. I know the right answer is .104g

The molar volume of a gas ($V_m$) at room temperature and standard pressure is $24$ liter.
That is, $1$ mole is $24$ liter.
How many moles does that make in $78\text{ cm}^3$?

 

[markosheehan]

Thanks.
I feel the answer is wrong. 78/24000 ×32 is what gives the answer at the back of the book. Should it not be multiplied by 16 though because in the balanced equation for the reaction it's a half mile of oxygen is formed.?

 

[I like Serena]

Thanks.
I feel the answer is wrong. 78/24000 ×32 is what gives the answer at the back of the book. Should it not be multiplied by 16 though because in the balanced equation for the reaction it's a half mile of oxygen is formed.?

Not sure what you mean. Was there a reaction involved?

We have 78/24000 moles of oxygen molecules ($O_2$).
Each molecule consists of 2 oxygen atoms, and an oxygen atom has mass 16u.
Therefore we multiply the moles with $2\times 16=32$.

 

[markosheehan]

sorry i forgot to mention it was calculate the mass of 78 cm3 of oxygen produced by the the composition of hydrogen peroxide.
h202 goes to h20 +.5o2
so does this not mean you would multiply it by 16 instead of 32