Calculate the mass of carbon dioxide that contains the same number of atoms of oxygen as present in #"1.8 g"# of water?

Answer 1

#"2.2 g CO"_2#

Start by calculating the number of moles of oxygen present in #"1.8 g"# of water.
As you know, one molecule of water contains two atoms of hydrogen and one atom of oxygen. This means that #1# mole of water contains #2# moles of hydrogen and #1# mole of oxygen.

Use the molar mass of water to convert the mass of water to moles.

#1.8 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.0color(red)(cancel(color(black)("g")))) = "0.10 moles H"_2"O"#

You can thus say that this sample contains

#0.10 color(red)(cancel(color(black)("moles H"_2"O"))) * "1 mole O"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.10 moles O"#
Now, for every #1# molecule of carbon dioxide, #"CO"_2#, you get #2# atoms of oxygen, so #1# mole of carbon dioxide will contain #2# moles of atoms of oxygen.

This means that the number of moles of oxygen present in your sample of water is enough to get you

#0.10 color(red)(cancel(color(black)("moles O"))) * "1 mole CO"_2/(2color(red)(cancel(color(black)("moless O")))) = "0.050 moles CO"_2" " " "color(darkorange)("(*)")#

Finally, to convert the number of moles of carbon dioxide to moles, use the molar mass of carbon dioxide.

#0.050 color(red)(cancel(color(black)("moles CO"_2))) * "44.0 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(darkgreen)(ul(color(black)("2.2 g")))#

The answer is rounded to two sig figs, the number of sig figs you have for the mass of water.

Notice that you didn't even have to convert the mass of water to atoms of oxygen because a mole of oxygen is simply a very large collection of atoms of oxygen, #6.022 * 10^(23)#, or #N_"A"#, to be precise #-># this is known as Avogadro's constant.

So if you want to double-check your answer, you can say that your sample of water contains

#0.10 color(red)(cancel(color(black)("moles O"))) * (N_"A" quad "atoms of O")/(1 color(red)(cancel(color(black)("mole O")))) = (0.10 * N_"A") quad "atoms of O"#
The same logic applies--you need #2# atoms of oxygen to get #1# molecule of carbon dioxide, so the number of atoms of oxygen present in your sample will get you
#(0.10 * N_"A") color(red)(cancel(color(black)("atoms O"))) * "1 molecule CO"_2/(2color(red)(cancel(color(black)("atoms O")))) = (0.050 * N_"A") quad "molecules CO"_2#

To convert the number of molecules of carbon dioxide to moles, use Avogadro's constant again.

#(0.050 * color(blue)(cancel(color(black)(N_ "A")))) color(red)(cancel(color(black)("molecules CO"_ 2))) * "1 mole CO"_ 2/(color(blue)(cancel(color(black)(N_ "A"))) color(red)(cancel(color(black)("molecules CO"_ 2)))) = "0.050 moles CO"_2#
This brings you right back to #color(darkorange)("(*)")#, so once again, the answer will be #"2.2 g"# of carbon dioxide.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To calculate the mass of carbon dioxide that contains the same number of oxygen atoms as 1.8 g of water, we first find the number of oxygen atoms in 1.8 g of water, then use this information to determine the mass of carbon dioxide.

  1. Calculate the number of moles of oxygen atoms in 1.8 g of water using the molar mass of water (H2O).
  2. Use the ratio of oxygen atoms in water (1:1) to find the number of moles of oxygen atoms.
  3. Use the Avogadro's number to find the number of oxygen atoms.
  4. Use the molar mass of carbon dioxide (CO2) to find the mass of carbon dioxide containing the same number of oxygen atoms.

Let's proceed with the calculations:

  1. Molar mass of water (H2O) = 18.015 g/mol Number of moles of oxygen atoms in 1.8 g of water = 1.8 g / 18.015 g/mol = 0.0999 mol (approximately 0.1 mol)

  2. Number of moles of oxygen atoms = 0.1 mol

  3. Number of oxygen atoms in 0.1 mol = 0.1 mol * 6.022 x 10^23 atoms/mol = 6.022 x 10^22 atoms

  4. Molar mass of carbon dioxide (CO2) = 44.01 g/mol Mass of carbon dioxide containing the same number of oxygen atoms = (6.022 x 10^22 atoms) * (44.01 g/mol / 2 atoms) ≈ 1.33 x 10^24 g

Therefore, the mass of carbon dioxide containing the same number of oxygen atoms as 1.8 g of water is approximately 1.33 x 10^24 grams.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7