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So funktioniert das nicht. Du kennst die Massen der Produkte. Daraus berechnest Du über die zugehörigen molaren Massen die entsprechenden Stoffmengen.
n(CO2) = m(CO2)/M(CO2) = 15,4 g/(44 g/mol) = 0,35 mol
n(H2O) = 7,2 g/(18 g/mol) = 0,40 mol
Nun interessieren uns allerdings die Stoffmengen von C und H, also den Bestandteilen des verbrannten Kohlenwasserstoffs. in einem mol CO2 steckt genau ein Mol C, da im Molekül pro CO2 ein Kohlenstoffatom vorhanden ist. Beim Wasser hat es pro Molekül jedoch 2 Wasserstoffe. Daher:
n(C) = n(CO2) = 0,35 mol
n(H) = 2n(H2O) = 0,80 mol
Damit haben wir bereits das richtige Stoffmengenverhältnis von C und H. Die vorläufige Verhältnisformel ist also:
C0,35H0,80
Das lässt man aber nicht so stehen, weil man möglichst ganzzahlige Indices hat. Deshalb suchen wir einen Faktor, der die Indizes ganzzahlig macht. Man kann jetzt zunnächst das Verhältnis der Indices bilden und findet H/C = 0,80/0,35 = 2,286. Das heißt, in dem gesuchten Kohlenwasserstoff ist die Stoffmenge an H um den Faktor 2,286 größere als der von C. Nehmen wir nun die allgemeine Formel für einen gesättigten Kohlenwasserstoff.
CnH2n+2
Damit kann man nun eine Bestimmungsgleichung erstellen.
2n+2 = 2.286n
n = 6,993 also gerundet 7. Damit haben wir den gesuchten Kohlenwasserstoff mit C7H16 (Heptan) gefunden. Und wenn man dazu die molare Masse berechnet, wird man sehen, dass diese 100 g/mol ist.
A hydrocarbon generally has the sum formula CxHy. The task is to find out x and y.
From the quantity ratio of CO2 and H2O that arises, you can first determine which relative The total formula of the hydrocarbon must have. So, like, n*CxHy. To do this, you need to consider what to burn: from each carbon atom present, a molecule of CO2 becomes and from two H atoms present, a molecule of H2 O becomes.
It was therefore very correct to calculate the figures: 2.1*10^23 molecules of CO2 and 2.4*10^23 molecules of H2O are formed. However, it is more pleasant to expect quantities of substances in mol: 0.35 mol of CO2 and 0.4 mol of H2 O are formed.
(Input: Does 2.1 x 10^23 mean that there are 2 CO2? No, that’s the absolute number of particles (molecules). By the way, if you convert from mol to absolute numbers, you shouldn’t write “mol” twice. 1 mol = 6*10^23… if you write: 6*10^23 mol, then the quasi mol2.
0.35 mol of CO2 are contained in 0.35 mol of C and 0.8 mol of H are contained in 0.4 mol of H2 O. So your hydrocarbon had a number ratio of 0.35 C to 0.8 H. Total numbers because there can only be whole numbers in total formulas. Have fun while recruiting, your result is x:y.
Your hydrocarbon therefore has the relative sum formula n*(CxHy). If you do not yet know the molar mass and do not remember the general formula for alkanes, n could also be=3 and the final formula would be 3*(CxHy)=C(3x)H(3y). Or n=4 and the final sum formula would be 4*(CxHy)=C(4x)H(4y).
Now to b). Here you are given the molar mass and can calculate n, and thus the final sum formula.
C has the molar mass of 12 g/mol; So Cx has the molar mass of… g/mol. Nevertheless, H has the molar mass 1 g/mol and Hy accordingly …. g/mol. Together, the parenthese of the relative sum formula … g/mol and then has to be n=… to finally come to the specified 100 g/mol. So this partial task is really simple.
But: she’s also a test if you worked correctly at a). If not, you would fail to come to the 100 g/mol with an integer formula (it happened to me earlier, which is why I had to recalculate).
As far as task c) is concerned, you have certainly discussed isomers of alkanes in class. Your searched hydrocarbon has 9.
To check if you have the right result, click here.
Thank you very much I understand!! I am glad that, in general, something at the a) I have partly calculated correctly
Once: On the Internet is “du”. And I’m glad I could get it over so you could start it!
The crucial point is to find an approach to the solution.
I didn’t really have the impression on your question that you knew what you had done these bills. So what they’re gonna do to you with the solution. What brought me to the question: was you aware that the result will not be “CH”?