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Glucose: C6 H12 O6

Carbon: 6 times 12 = 72

Hydrogen: 12 times 1 (actually 1.008) = 12

Oxygen: 6 times 16 = 96

All together that makes the molecular weight of Glucose is 180

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Q: If the atomic weight of carbon is twelve and the atomic weight of oxygen is sixteen the molecular weight of glucose is?
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If the atomic weight of carbon is twelve and the atomic weight of oxygen is sixteen the molecular weight of glucose expressed in grams is?

The question lacks sufficent information regarding quantity, and atomic weight is not expressed in grams. Are you asking how many grams per mole?


A glucose molecule contains carbon hydrogen and oxygen in the ratio 6 to 12 to 6 What is the molecular formula for glucose?

Molecular formulas can not be determined from information about atomic ratios only. If the ratios stated are atomic ratios, then the empirical formula of the compound is CH2O. If the questioner meant, "A glucose molecule contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms ...", the molecular formula is C6H12O6.


Is carbon a molecular or atomic element?

atomic


How do you get the molecular mass of glucose?

A molecule of glucose contains 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms. Their atomic masses are 12, 1 and 16 respectively. Therefore the atomic mass of glucose is 6x12+12x1+6x16=180 grams per mole.


What is the number of carbon atoms in a molecule of glucose?

C6 H12 O6 being the molecular formula of glucose, the carbon atoms are 6 in one molecule of glucose.


Molecular formula of glucose?

C6H12O6. 6 atoms of carbon, 12 atoms of hydrogen, 6 atoms of oxygen.


What are the gram molecular weights ofC6H8O7?

gram molecular weight of C6H8O7=(atomic weigth of carbon X 6)+(atomic weigth of hydrogen X 8+(atomic weigth of oxygen X 7) grams


What percentage of each glucose molecule is made up of carbon?

* Percentage mass = atomic mass of required element/ total mass of compound. * Glucose is represented by the formula C6H12O6. The relative molecular mass (Mr)= (6 x 12) + (12 x 1) + (6 x 16) = 180. # Let's start with carbon: one atom of carbon has an atomic mass of 12, but there are 6 atoms of carbon in an molecule of glucose. 6 x 12 = 72. Divide this by 180 and multiply by 100 to give 40%. # Next, we have hydrogen. Each atom has an atomic mass of 1. Multiply by 12 to give 12. Then divide by 180 and multiply by 100, to give 6.6% # Finally, there's oxygen. One atom of oxygen has an atomic mass of 16, so multiply 16 by 6, which gives 96. Divide this by 180, then multiply by 100 to give 53.3%. * To check that these % masses are correct, add the percentages to ensure they make 100: 40 + 6.6 + 53.3 = 99.9%, which is close enough!


Is molecular formula a element?

Technically no, molecular formula tells you how many atoms of each element compose a molecule. For example O: oxygen C: Carbon H: hydrogen Carbon dioxides molecular formula is CO2 Atmospheric oxygen is O2 Glucose's molecular formula is C6H12O6


Is carbon oxygen a compound or element?

Carbon monoxide (CO) and carbon dioxide (CO2) are molecular compounds containing the elements carbon and oxygen. There is no compound named carbon oxygen.Added:Carbon is an element, atomic symbol C.Oxygen is an element, di-atomic molecule with symbol O2.Carbon dioxide is a compound with molecular formula CO2, where the two elements (C and O2) are bonded together.Carbon monoxide is a compound, (very different from carbon dioxide), with molecular formula CO, where the two elements (C and O2) are bonded together in a different way.


What is the molecular weight of carbon?

The molecular weight of carbon is 12.0107 g/mol, or grams per molecule. This is calculated by dividing the total weight of the atom by the weight of all the atomic mass units and multiplying by 100.


How many different elements are in a molecule of glucose?

A molecule of glucose has three different elements viz., carbon, oxygen and hydrogen. The molecular formula of glucose is C6H12O6. There is actually 24 different elements. There are 6 Carbon elements,12 Hydrogen elements, and 6 Oxygen elements.