Calculating the mass and number of atoms of a structural I beam.

AI Thread Summary
To calculate the mass and number of atoms in a structural I beam made of iron, the volume must be determined accurately, considering both the solid and hollow sections. The density of iron is 7560 kg/m^3, and the atomic weight is 55.85 g/mol. After calculating the volume using the dimensions provided, the mass was found to be approximately 198.949 kg, which rounds to 200 kg. The method suggested involved calculating the volume of the beam's components separately for accuracy. The final mass calculation was confirmed to be correct, demonstrating the importance of precise volume calculations in such problems.
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Homework Statement


A structural I beam is made of iron> A view of its cross-section and its dimensions is shown in the figure, where d = 2 cm, w= 31 cm, h= 28 cm and the length (not shown) of the beam is L= 1.53 m.
fndyzp.jpg

The density of iron is 7560 kg/m^3. The atomic weight of iron is 55.85 g/mol and Avagadro number is Na= 6.02214e23/mol.

What is the mass of a section 1.53 m long? (in kg)
How many atoms are there in this section? (in units of atoms)

The Attempt at a Solution



I calculated the volume of the I beam ( (h*w*l)-(the volume of the empty space))
and then I multiply it by the density of iron to get the mass.

but the answer was wrong, thus i I tried to calculate the number of atoms using Avagadro's number the answer will be wrong as well...

I hope you help me calculating the mass...

Thank you
 
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Your method is fine, you probably made a math error in calculating the dimensions of the empty space, or perhaps you forgot to convert centimeters to meters. You should show your work so that we can point out the errors. You might want to calculate the volume in another way, by looking at the volume of the three separate parts of the beam (top, bot, and middle pieces) and add them up, as a check on your work.
 
Thanks for your reply PhanthomJay

I used the method that you suggested for calculating the volume

((0.31)*(0.02)*(1.53))*2 + (0.24)*(0.02)*(1.53) = 0.026316 m^2

thus, mass= 0.026316 m^2 * density of iron= 198.949 Kg
I rounded it to 1 significant figures and I got 2.0e2 kg

Good news ... answer is correct

thanks a lot for the help
 
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