The Elements of Mars
- Page ID
- 49958
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Work in progress. Rhetue 15:38, 12 April 2011 (UTC)
Before liquid water was found on Mars, astronomers had pointed to the presence of ferric oxide, responsible for Mars' red color, as an indicator that water once flowed on its surface. Ferric oxide, also known as a "rust", forms when iron metal is left in standing water and is oxidized. The Mars Phoenix Lander, which touched down on the surface of Mars in 2008, carried instruments that analyzed the soil of Mars and found ice samples in the soil mixed in with ferric oxide dust[1]. Phoenix used a sophisticated "bake and sniff" oven that analyzed vaporized samples and identified the compounds it contained[2]. The data from volatile compounds released are sent back to Earth and analyzed.
One of the simpler ways to distinguish chemicals is by their molecular weights, which depend on individual atomic weights. An atom of each element has a specific mass ratio to every other element, with hydrogen's mass set at 1. Thus nitrogen’s atomic weight of 14.0067 tells us that a nitrogen atom has about 14 times the mass of a hydrogen atom. Often, these values are expressed with units of atomic mass units, or "amu", though as ratios, they are technically unitless.
Molecular weights are calculated by summing the atomic weights of all atoms in the molecule. The Table of Atomic Weights lists atomic weights for all elements. Atomic weights remain constant throughout reactions, regardless of the environment.
Example \(\PageIndex{1}\): Molecular Weights
Calculate the molecular weights of
- water (H2O) and
- ferric oxide (Fe2O3).
- Obtain the mass ratio of ferric oxide to water.
Answer
The molecular weight of each compound is obtained by summing the atomic weights of their constituent atoms.
a) Molecular weight of H2O:
\(\text{2}\times \text{(atomic weight of H)} + \text{(atomic weight of O)} = \text{(2 x 1)} + \text{16} = \text{18 amu}\)
b) Molecular weight of Fe2O3:
\(\text{2} \times \text{(atomic weight of Fe)} + \text{3} \times \text{(atomic weight of O)} = \text{159.7 amu}\)
c) Mass ratio of ferric oxide to water =
\(\dfrac{\text{159.7 amu}}{\text{18 amu}} = \text{ 9 : 1 ratio}\)
The mass ratio is useful by roughly determining how much heavier a molecule is compared to another.
Example \(\PageIndex{2}\): Mass Percentage
A Martian soil sample weighing 25 g was collected by Phoenix and assumed to be comprised of only Fe2O3 and H2O. It was analyzed and found to be 24% elemental Fe, but instruments were unable to accurately measure the percentage of hydrogen and oxygen in the sample. Can the % mass of Fe2O3 be obtained, and if so, what is its value?
Answer
The % mass of Fe2O3 can be obtained from given data, by making the key realizations that the mass of Fe in the sample is contributed only by Fe2O3, and that for a given amount of Fe, a stoichiometric amount of O must also be present. Because we cannot find the amount of hydrogen in the sample, we are initially unable to determine the mass of water. However, we can obtain an actual mass of iron in the sample by using elemental % mass, and from that value, the molar amount of iron in the sample.
Because Fe2O3 is a compound, we know that a certain molar ratio of oxygen will always be present with the iron, and use the ratio to obtain the moles of oxygen present from Fe2O3. The combined masses of the given iron and calculated oxygen account for the total mass of Fe2O3. This value divided by the mass of the entire sample will give the % mass of Fe2O3.
Let's go through this explanation with calculations.
1) Find the actual mass of Fe in the sample by multiplying the mass of the sample by the given % mass of Fe.
\(\text{0.24} \times \text{25 g} =\text{ 6 g Fe in the sample}\)
2) Convert this value to moles, using molar mass.
\(\text{6 g} \times \dfrac{\text{1 mol}}{\text{55.845 g}} = \text{0.1074 mol Fe}\)
3) We converted to moles in order to use the stoichiometric ratio of Fe to O.
\(\text{0.1074 mol Fe} \times \dfrac{\text{3 mol O}}{\text{2 mol Fe}} = \text{0.1612 mol O from ferric oxide}\)
4) Mole of O are converted to
\(\text{0.1612 mol O} \times \dfrac{\text{16g}}{\text{1 mol}} = \text{2.58 g O from ferric oxide}\)
5) Mass % of ferric oxide is obtained by dividing the combined masses of Fe and O from Fe2O3 by the mass of the sample.
\(\dfrac{\text{6 + 2.58 g}}{\text{25 g}} = \text{34/32 percent ferric oxide}\)
Note that we cannot assume that this accounts for all the oxygen present in the sample
From ChemPRIME: 2.5: Atomic Weights
References
- news.nationalgeographic.com/n...s-water_2.html
- http://www.nasa.gov/mission_pages/phoenix/news/phoenix-20080611.html
Contributors and Attributions
Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.