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magnesium carbonate hydroxide pentahydrate

Input interpretation

magnesium carbonate hydroxide pentahydrate
magnesium carbonate hydroxide pentahydrate

Basic properties

molar mass | 192.4 g/mol formula | CH_12MgO_9 empirical formula | O_9C_Mg_H_12 SMILES identifier | C(=O)(O)[O-].[Mg+2].O.O.O.O.O.[OH-] InChI identifier | InChI=1/CH2O3.Mg.6H2O/c2-1(3)4;;;;;;;/h(H2, 2, 3, 4);;6*1H2/q;+2;;;;;;/p-2/fCHO3.Mg.5H2O.HO/h2H;;;;;;;1h/q-1;m;;;;;;-1 InChI key | PWRJGELXVCNTCM-UHFFFAOYSA-L
molar mass | 192.4 g/mol formula | CH_12MgO_9 empirical formula | O_9C_Mg_H_12 SMILES identifier | C(=O)(O)[O-].[Mg+2].O.O.O.O.O.[OH-] InChI identifier | InChI=1/CH2O3.Mg.6H2O/c2-1(3)4;;;;;;;/h(H2, 2, 3, 4);;6*1H2/q;+2;;;;;;/p-2/fCHO3.Mg.5H2O.HO/h2H;;;;;;;1h/q-1;m;;;;;;-1 InChI key | PWRJGELXVCNTCM-UHFFFAOYSA-L

Structure diagram

Structure diagram
Structure diagram

Quantitative molecular descriptors

longest chain length | 3 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 0 atoms H-bond acceptor count | 9 atoms H-bond donor count | 6 atoms
longest chain length | 3 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 0 atoms H-bond acceptor count | 9 atoms H-bond donor count | 6 atoms

Elemental composition

Find the elemental composition for magnesium carbonate hydroxide pentahydrate in terms of the atom and mass percents: atom percent = N_i/N_atoms × 100% mass percent = (N_im_i)/m × 100% Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i, m_i, N_atoms and m using these items. • Finally, compute the percents and check the results. Write the chemical formula: CH_12MgO_9 Use the chemical formula to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms, per molecule:  | number of atoms  O (oxygen) | 9  C (carbon) | 1  Mg (magnesium) | 1  H (hydrogen) | 12  N_atoms = 9 + 1 + 1 + 12 = 23 Divide each N_i by N_atoms to calculate atom fractions. Then use the property that atom fractions must sum to one to check the work:  | number of atoms | atom fraction  O (oxygen) | 9 | 9/23  C (carbon) | 1 | 1/23  Mg (magnesium) | 1 | 1/23  H (hydrogen) | 12 | 12/23 Check: 9/23 + 1/23 + 1/23 + 12/23 = 1 Compute atom percents using the atom fractions:  | number of atoms | atom percent  O (oxygen) | 9 | 9/23 × 100% = 39.1%  C (carbon) | 1 | 1/23 × 100% = 4.35%  Mg (magnesium) | 1 | 1/23 × 100% = 4.35%  H (hydrogen) | 12 | 12/23 × 100% = 52.2% Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table:  | number of atoms | atom percent | atomic mass/u  O (oxygen) | 9 | 39.1% | 15.999  C (carbon) | 1 | 4.35% | 12.011  Mg (magnesium) | 1 | 4.35% | 24.305  H (hydrogen) | 12 | 52.2% | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m:  | number of atoms | atom percent | atomic mass/u | mass/u  O (oxygen) | 9 | 39.1% | 15.999 | 9 × 15.999 = 143.991  C (carbon) | 1 | 4.35% | 12.011 | 1 × 12.011 = 12.011  Mg (magnesium) | 1 | 4.35% | 24.305 | 1 × 24.305 = 24.305  H (hydrogen) | 12 | 52.2% | 1.008 | 12 × 1.008 = 12.096  m = 143.991 u + 12.011 u + 24.305 u + 12.096 u = 192.403 u Divide the mass for each element by m to calculate mass fractions. Then use the property that mass fractions must sum to one to check the work:  | number of atoms | atom percent | mass fraction  O (oxygen) | 9 | 39.1% | 143.991/192.403  C (carbon) | 1 | 4.35% | 12.011/192.403  Mg (magnesium) | 1 | 4.35% | 24.305/192.403  H (hydrogen) | 12 | 52.2% | 12.096/192.403 Check: 143.991/192.403 + 12.011/192.403 + 24.305/192.403 + 12.096/192.403 = 1 Compute mass percents using the mass fractions: Answer: |   | | number of atoms | atom percent | mass percent  O (oxygen) | 9 | 39.1% | 143.991/192.403 × 100% = 74.84%  C (carbon) | 1 | 4.35% | 12.011/192.403 × 100% = 6.243%  Mg (magnesium) | 1 | 4.35% | 24.305/192.403 × 100% = 12.63%  H (hydrogen) | 12 | 52.2% | 12.096/192.403 × 100% = 6.287%
Find the elemental composition for magnesium carbonate hydroxide pentahydrate in terms of the atom and mass percents: atom percent = N_i/N_atoms × 100% mass percent = (N_im_i)/m × 100% Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i, m_i, N_atoms and m using these items. • Finally, compute the percents and check the results. Write the chemical formula: CH_12MgO_9 Use the chemical formula to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms, per molecule: | number of atoms O (oxygen) | 9 C (carbon) | 1 Mg (magnesium) | 1 H (hydrogen) | 12 N_atoms = 9 + 1 + 1 + 12 = 23 Divide each N_i by N_atoms to calculate atom fractions. Then use the property that atom fractions must sum to one to check the work: | number of atoms | atom fraction O (oxygen) | 9 | 9/23 C (carbon) | 1 | 1/23 Mg (magnesium) | 1 | 1/23 H (hydrogen) | 12 | 12/23 Check: 9/23 + 1/23 + 1/23 + 12/23 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 9 | 9/23 × 100% = 39.1% C (carbon) | 1 | 1/23 × 100% = 4.35% Mg (magnesium) | 1 | 1/23 × 100% = 4.35% H (hydrogen) | 12 | 12/23 × 100% = 52.2% Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | number of atoms | atom percent | atomic mass/u O (oxygen) | 9 | 39.1% | 15.999 C (carbon) | 1 | 4.35% | 12.011 Mg (magnesium) | 1 | 4.35% | 24.305 H (hydrogen) | 12 | 52.2% | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: | number of atoms | atom percent | atomic mass/u | mass/u O (oxygen) | 9 | 39.1% | 15.999 | 9 × 15.999 = 143.991 C (carbon) | 1 | 4.35% | 12.011 | 1 × 12.011 = 12.011 Mg (magnesium) | 1 | 4.35% | 24.305 | 1 × 24.305 = 24.305 H (hydrogen) | 12 | 52.2% | 1.008 | 12 × 1.008 = 12.096 m = 143.991 u + 12.011 u + 24.305 u + 12.096 u = 192.403 u Divide the mass for each element by m to calculate mass fractions. Then use the property that mass fractions must sum to one to check the work: | number of atoms | atom percent | mass fraction O (oxygen) | 9 | 39.1% | 143.991/192.403 C (carbon) | 1 | 4.35% | 12.011/192.403 Mg (magnesium) | 1 | 4.35% | 24.305/192.403 H (hydrogen) | 12 | 52.2% | 12.096/192.403 Check: 143.991/192.403 + 12.011/192.403 + 24.305/192.403 + 12.096/192.403 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 9 | 39.1% | 143.991/192.403 × 100% = 74.84% C (carbon) | 1 | 4.35% | 12.011/192.403 × 100% = 6.243% Mg (magnesium) | 1 | 4.35% | 24.305/192.403 × 100% = 12.63% H (hydrogen) | 12 | 52.2% | 12.096/192.403 × 100% = 6.287%

Elemental oxidation states

The first step in finding the oxidation states (or oxidation numbers) in magnesium carbonate hydroxide pentahydrate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge:  In magnesium carbonate hydroxide pentahydrate hydrogen is not bonded to a metal with lower electronegativity, so it will have an oxidation state of +1. Any element bonded to hydrogen gains the bonding electrons, decreasing their oxidation state by 1 for every bond:  With hydrogen out of the way, look at the remaining bonds. There are 3 carbon-oxygen bonds. For each of these bonds, assign the bonding electrons to the most electronegative element.  First examine the carbon-oxygen bonds: element | electronegativity (Pauling scale) |  C | 2.55 |  O | 3.44 |   | |  Since oxygen is more electronegative than carbon, the electrons in these bonds will go to oxygen. Decrease the oxidation number for oxygen in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for carbon accordingly:  Now summarize the results: Answer: |   | oxidation state | element | count  -2 | O (oxygen) | 9  +1 | H (hydrogen) | 12  +2 | Mg (magnesium) | 1  +4 | C (carbon) | 1
The first step in finding the oxidation states (or oxidation numbers) in magnesium carbonate hydroxide pentahydrate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In magnesium carbonate hydroxide pentahydrate hydrogen is not bonded to a metal with lower electronegativity, so it will have an oxidation state of +1. Any element bonded to hydrogen gains the bonding electrons, decreasing their oxidation state by 1 for every bond: With hydrogen out of the way, look at the remaining bonds. There are 3 carbon-oxygen bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-oxygen bonds: element | electronegativity (Pauling scale) | C | 2.55 | O | 3.44 | | | Since oxygen is more electronegative than carbon, the electrons in these bonds will go to oxygen. Decrease the oxidation number for oxygen in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for carbon accordingly: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 9 +1 | H (hydrogen) | 12 +2 | Mg (magnesium) | 1 +4 | C (carbon) | 1

Orbital hybridization

hybridization | element | count  sp^2 | C (carbon) | 1  | O (oxygen) | 3  sp^3 | O (oxygen) | 6
hybridization | element | count sp^2 | C (carbon) | 1 | O (oxygen) | 3 sp^3 | O (oxygen) | 6

Structure diagram

Orbital hybridization Structure diagram
Orbital hybridization Structure diagram

Topological indices

vertex count | 23 edge count | 15 Schultz index |  Wiener index |  Hosoya index |  Balaban index |
vertex count | 23 edge count | 15 Schultz index | Wiener index | Hosoya index | Balaban index |